Series of the day

The sum $\displaystyle \sum\frac{1}{(n+1)^pn^q}$

Today we prove the following result involving the Riemann zeta function $$\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} = (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + \binom{m+r+1}{r}\right]$$ In the proof requires we use some properties of the digamma and the polygamma functions.

Proof:

The difference between two digamma functions may be expressed as $$(a-b)\sum_{j=0}^{\infty} \frac{1}{(j+a)(j+b)} = \psi(a)-\psi(b)\tag{1}$$ if we make change the of variable $j=n-1$ $$\sum_{n=1}^{\infty} \frac{1}{(n-1+a)(n-1+b)} = \frac{\psi(a)-\psi(b)}{a-b}$$ Recall the Leibnitz rule for higher-order derivatives: $$\frac{d^n}{dx^n} f(x)g(x) = \sum_{j=1}^n \binom{n}{j} f^{(n-j)}(x)g^{(n)}(x)$$ Differentiating $(1)$ both sides with respect to $a$, $r$ times: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{(-1)^r r!}{(n-1+a)^{r+1}(n-1+b)} &=& \sum_{j=0}^{r}\binom{r}{j}\left(\frac{d^{n-j}}{dx^{n-j}} \frac{1}{(a-b)}\right) \left(\frac{d^{j}}{da^{j} } \psi(a) - \psi(b)\right)\\ \\ & = &\sum_{j=0}^{r}\binom{r}{j} \left(\frac{(-1)^{r-j} (r-j) !}{(a-b)^{r-j+1}}\right)\left(\frac{d^{j}}{da^{j} } \psi(a) - \psi(b)\right)\\ &=& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(a)}{(a-b)^{r-j+1}} + (-1)^{r} r!\frac{\psi(a)-\psi(b)}{(a-b)^{r+1}} \end{eqnarray*}$$ Where $\psi^{(n)}(x)$ is the polygamma function. This function has the following expansion: $$\psi^{(n)}(x) = n! \sum_{j=0}^{\infty} \left(\frac{-1}{j+x}\right)^{n+1} \tag{2}$$ from this expansion we can derive the following relation: $$\psi^{(n)}(1) = n! (-1)^{n+1}\zeta(n+1) \tag{3}$$ This function also satisfies the following recurrence formula: $$\psi^{(n)}(x+1) =\psi^{(n)}(x) - \frac{n!}{(-x)^{n+1}} \tag{4}$$ So, if we let $a\to 2$ then $\psi(2) =1 -\gamma$: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{ (-1)^rr!}{(n+1)^{r+1}(n-1+b)} & =& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(2)}{(2-b)^{r-j+1}} + (-1)^{r} r!\frac{1-\gamma-\psi(b)}{(a-b)^{r+1}}\\ & =& \sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !\psi^{(j)}(1)}{(2-b)^{r-j+1}} -\sum_{j=1}^{r}\binom{r}{j}\frac{(-1)^{r-j} (r-j) !j!}{(-1)^{j+1}(2-b)^{r-j+1}} + (-1)^{r} r!\frac{1-\gamma-\psi(b)}{(a-b)^{r+1}}\quad \textrm{ from } (4)\\ \end{eqnarray*}$$ After rearranging the expression: $$\begin{eqnarray*} \sum_{n=1}^{\infty} \frac{ 1}{(n+1)^{r+1}(n-1+b)} &= & -\sum_{j=1}^{r}\frac{ \zeta(j+1)}{(2-b)^{r-j+1}} +\sum_{j=1}^{r}\frac{ 1}{(2-b)^{r-j+1}}+ \frac{1-\gamma-\psi(b)}{(2-b)^{r+1}} \quad \textrm{ from } (3) \end{eqnarray*}$$ If we now differentiate both sides with respect to $b$, $m$ times: $$\begin{eqnarray*}\sum_{n=1}^{\infty} \frac{ (-1)^m m!}{(n+1)^{r+1}(n-1+b)^{m+1}} & = & -\sum_{j=1}^{r}\frac{ (r-j+m)!\zeta(j+1)}{(r-j)!(2-b)^{r-j+m+2}} + \sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!(2-b)^{r-j+1}} \\ &-& \sum_{k=1}^{m}\binom{m}{k}\frac{ (r+m-k) !\psi^{k}(b)}{r! (2-b)^{r+m-k+2}} + \frac{(r+m)!}{r!} \frac{(1-\gamma - \psi(b))}{(2-b)^{r+m+2}}\\ \end{eqnarray*}$$ If we let $b\to 1$ then $\psi(1) = -\gamma$ and using again the relationship between $\psi^{(n)}(1)$ and $\zeta(n+1)$ $$\begin{eqnarray*}\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} & = & -(-1)^m\sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!m!} \zeta(j+1) + (-1)^m\sum_{j=1}^{r}\frac{ (r-j+m)!}{(r-j)!m!}\\ &+& (-1)^m\sum_{k=1}^{m}\frac{(r+m-k) !}{(m-k)!r!}(-1)^k\zeta(k+1) + (-1)^m \frac{(r+m)!}{r! m!} \\ & = & -(-1)^m\sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + (-1)^m\sum_{j=1}^{r}\binom{ r-j+m}{m}\\ &+& (-1)^m\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) + (-1)^m\binom{r+m}{m} \end{eqnarray*}$$ We can prove by induction that $$\sum_{j=1}^{r}\binom{ r-j+m}{m} = \binom{r+m}{m+1}$$ Base case, $r=1$ $$\sum_{j=1}^{1}\binom{ 1-j+m}{m} = \binom{ m}{m} = \binom{ m+1}{m+1}$$ Inductive step, suppose the formula is valid for $r = p$ $$\sum_{j=1}^{p}\binom{ p-j+m}{m} = \binom{p+m}{m+1}$$ Then, for $r=p+1$: $$\begin{eqnarray*} \sum_{j=1}^{p+1}\binom{ p+1-j+m}{m} & =& \sum_{k=0}^{p}\binom{ p-k+m}{m} \\ & =& \binom{p+m}{m} + \sum_{k=1}^{p}\binom{ p-k+m}{m}\\ & = & \binom{p+m}{m} + \binom{p+m}{m+1}\\ & =& \binom{p+1+m}{m+1 } \end{eqnarray*}$$ Therefore: $$\begin{eqnarray*} \sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} &=& (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) +\binom{r+m}{m+1} + \binom{r+m}{m} \right] \\ &=& (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) +\binom{r+m+1}{r} \right] \end{eqnarray*}$$ Hence $$\boxed{\sum_{n=0}^{\infty} \frac{ 1}{(n+1)^{r+1}n^{m+1}} = (-1)^m\left[\sum_{k=1}^{m}\binom{r+m-k}{r}(-1)^k\zeta(k+1) - \sum_{j=1}^{r}\binom{ r-j+m}{m} \zeta(j+1) + \binom{m+r+1}{r}\right]}$$

Application of Ramanujan's master theorem III

Strange integral involving the Euler-Mascheroni constant $\gamma$

Today we show the proof of this strange integral posted by @integralsbot $$\int_{0}^{\infty} \frac{\cos(x^2)-\cos(x)}{x}dx = \frac{\gamma}{2}$$ In the way to find the solution we also find the following Mellin transform $$\int_{0}^{\infty}w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw = \Gamma(s)\cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)} \quad 0\lt s \lt 1$$ The proof relies on the powerful Ramanujan's master theorem.

Proof:

Recall the Ramanujan's master theorem:

If a complex-valued function $f(x)$ has an expansion of the form $$f(x)=\sum_{k=0}^\infty \frac{\,\varphi(k)\,}{k!}(-x)^k$$ then the Mellin transform of $f(x)$ is given by $$\int_0^\infty x^{s-1} f(x) dx = \Gamma(s)\,\varphi(-s) \tag{1}$$ Back to the integral: $$\int_{0}^{\infty} \frac{\cos(x^2)-\cos(x)}{x}dx = \frac{1}{2} \int_{0}^{\infty} \frac{\cos(w)-\cos(\sqrt{w})}{w}dw \quad (w \mapsto x^2)$$ Consider the following Mellin transform $$\mathscr{M}\left\{ \cos(w)-\cos(\sqrt{w}) \right\} = \int_{0}^{\infty}w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw$$ Note that we can expand the cosine function in the following way: $$\begin{align*} \cos(w) = \Re\left(e^{iw}\right) =& \Re\left(\sum_{n=0}^{\infty} \frac{i^nw^n}{n!}\right) \\ =& \Re\left(\sum_{n=0}^{\infty} \frac{(-i^n)(-w^n)}{n!}\right) \\ =& \sum_{n=0}^{\infty} \frac{\Re((-i^n))(-w^n)}{n!} \\ =& \sum_{n=0}^{\infty} \frac{\cos\left(\frac{\pi n}{2} \right) (-w^n)}{n!} \end{align*}$$ and we can exapnd $\cos(\sqrt{w})$ in the traditional way: $$\begin{align*} \cos(\sqrt{w}) = \sum_{n=0}^{\infty} \frac{(-w)^n}{(2n)!} \end{align*}$$ Therefore $$\begin{align*} \cos(w)-\cos(\sqrt{w}) =& \sum_{n=0}^{\infty} \frac{\cos\left(\frac{\pi n}{2} \right) (-w^n)}{n!} - \sum_{n=0}^{\infty} \frac{(-x)^n}{(2n)!}\\ =& \sum_{n=0}^{\infty}\left[ \frac{\cos\left(\frac{\pi n}{2} \right) (-w^n)}{n!} - \frac{(-w)^n}{(2n)!}\right]\\ =& \sum_{n=0}^{\infty}\underbrace{\left[ \cos\left(\frac{\pi n}{2} \right) - \frac{n!}{(2n)!}\right]}_{\varphi(n)}\frac{(-w^n)}{n!}\ \end{align*}$$ Applying the Ramanujan's master theorem : $$\begin{align*} \int_{0}^{\infty}w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw = &\Gamma(s) \varphi(-s) \quad \textrm{from (1)}\\ =& \Gamma(s)\left[ \cos\left(-\frac{\pi s}{2} \right) - \frac{(-s)!}{(-2s)!}\right]\\ =& \Gamma(s)\left[ \cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(1-s)}{\Gamma(1-2s)}\right]\\ =& \Gamma(s)\cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\ \end{align*}$$ Hence $$\int_{0}^{\infty}w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw = \Gamma(s)\cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}$$ We can use this Mellin transform to calculate our integral Taking the limit as $s \to 0+$ $$\begin{align*} \int_{0}^{\infty} \frac{\cos(w) -\cos(\sqrt{w}) }{w}dw =& \int_{0}^{\infty} \lim_{s \to 0+} w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw\\ & \lim_{s \to 0+} \int_{0}^{\infty}w^{s-1} \left(\cos(w) -\cos(\sqrt{w})\right) dw\\ =& \lim_{s \to 0+} \Gamma(s)\left[ \cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(1-s)}{\Gamma(1-2s)}\right]\\ =& \lim_{s \to 0+} \Gamma(s+1) \lim_{s \to 0+} \frac{\left[ \cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(1-s)}{\Gamma(1-2s)}\right]}{s}\\ =& \frac{ \lim_{s \to 0+} \frac{d}{ds} \left[\cos\left(\frac{\pi s}{2} \right) - \frac{\Gamma(1-s)}{\Gamma(1-2s)}\right]}{ \lim_{s \to 0+} \frac{d}{ds} s }\\ =& \lim_{s \to 0+}-\frac{1}{2}\sin\left(\frac{\pi s}{2}\right)+ \frac{\Gamma(1-s)}{\Gamma(1-2s)} \left(\psi(1-s)-2\psi(1-2s)\right)\\ =& \lim_{s \to 0+} \psi(1-s)-2\psi(1-2s)\\ =& -\psi(1)\\ =& \gamma \end{align*}$$ Hence $$\int_{0}^{\infty} \frac{\cos(w) -\cos(\sqrt{w}) }{w}dw = \gamma$$ Therefore we can conclude $$\int_{0}^{\infty} \frac{\cos(x^2)-\cos(x)}{x}dx = \frac{1}{2} \int_{0}^{\infty} \frac{\cos(w)-\cos(\sqrt{w})}{w}dw = \frac{\gamma}{2}$$ $$\boxed{ \int_{0}^{\infty} \frac{\cos(x^2)-\cos(x)}{x}dx = \frac{\gamma}{2} }$$

Generalized hypergeometric functions XI

Nice hypergeometric series involving the silver ratio $\delta_S$

Today we show the proof of this result involving hypergeometric series and the silver ratio $\delta_S= 1+\sqrt{2}$ proposed by @mamekebi $${}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}$$ The proof relies on some properties of the rising factorial and some identities of the dilogarithm function.

Proof

Recall the recursion formula for the argument and the degree of the rising factorial: $$(x)_{n+1} = (n+x)(x)_{n} = x(x+1)_{n}\quad \tag{1}$$ and the duplication formula $$(x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_n\left(\frac{x+1}{2}\right)_n \tag{2}$$ Hence $$\begin{align*} {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = &\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}\left(\frac{1}{2}\right)_{j}}{\left(\frac{3}{2}\right)_{j}\left(\frac{3}{2}\right)_{j}} \frac{(-1)^j}{j!}\\ =&\frac{1}{2}\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(j+\frac{1}{2}\right)} \frac{(-1)^j}{j!} \quad \textrm{from (1)}\\ =&\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(2j+1\right)} \frac{(-1)^j}{j!}\\ \end{align*}$$ We previously showed the proof of this series expansion: $$\begin{align*}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} =& \sum_{j=0}^{\infty} \frac{2^{2j}}{\binom{2j}{j}(2j+1)}(-1)^jx^{2j} = \sum_{j=0}^{\infty} \frac{2^{2j}j!^2}{(2j)!(2j+1)}(-1)^jx^{2j} \\ =& \sum_{j=0}^{\infty} \frac{2^{2j}(1)_{j}(1)_{j}}{(1)_{2j}(2j+1)}(-1)^jx^{2j} \quad \left( (1)_{j} = j!\right)\\ =& \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}}\frac{(-1)^jx^{2j}}{j!} \quad \textrm{ from (1) and (2)} \end{align*}$$ Integrating from $0$ to $1$ $$\int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}(2j+1)}\frac{(-1)^j}{j!} ={}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right]$$ $$\begin{align*} I = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = &\int_{0}^{1}\frac{\ln\left(\sqrt{x^2+1}+x\right)}{x\sqrt{1+x^2}} dx\\ =& 2\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{(w+1)(w-1)}dw \quad \left(w \mapsto \sqrt{x^2+1}+x\right) \\ =& -\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{1-w}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw\\ =& \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw \quad \left(t\mapsto 1-w\right) \\ \stackrel{IBP}{=} & \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{-1}\frac{\ln(1-w)}{w}dw\\ =& -\int_{-\sqrt{2}}^{0}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{0}\frac{\ln(1-w)}{w}dw+\int_{-1}^{0}\frac{\ln(1-w)}{w}dw\\ =& -\operatorname{Li}_{2}(-\sqrt{2})- \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\operatorname{Li}_{2}(-1-\sqrt{2})+\operatorname{Li}_{2}(-1) \end{align*}$$ From the identity $$\operatorname{Li}_2(-z)-\operatorname{Li}_2(1-z)+\frac{1}{2}\operatorname{Li}_2(1-z^2)=-\frac{{\pi}^2}{12}-\ln z \cdot \ln(z+1)$$ If we put $z = \sqrt{2}$ $$\operatorname{Li}_2(-\sqrt{2})=\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{2}\operatorname{Li}_2(-1)-\frac{{\pi}^2}{12}-\ln \sqrt{2} \cdot \ln(\sqrt{2}+1) \quad \quad (3)$$ From the identity $$\operatorname{Li}_2(z) +\operatorname{Li}_2\left(\frac{1}{z}\right) = - \frac{\pi^2}{6} - \frac{\ln^2(-z)}{2}$$ If we put $z = 1-\sqrt{2}$

Therefore $$\operatorname{Li}_2\left(-1-\sqrt{2}\right) = -\operatorname{Li}_2(1-\sqrt{2}) - \frac{\pi^2}{6} - \frac{\ln^2(\sqrt{2}-1)}{2} \quad \quad (4)$$ Using the fact that $$\operatorname{Li}_{2}(-1)= -\frac{\pi^2}{12}$$ and from $(3)$ and $(4)$ we have: $$\begin{align*} I =& \frac{\pi^2}{8} +\ln \sqrt{2} \cdot \ln(\sqrt{2}+1)- \ln(1+\sqrt{2})\ln(2+\sqrt{2}) + \frac{\ln^2(\sqrt{2}-1)}{2}\\ =& \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2} \end{align*}$$ $$\boxed{ {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}}$$

Generalized hypergeometric functions X

Series involving the binomial coefficient

Today we show the proof of this series proposed by @qq3023625451 $$\sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}$$ In the proof we use some properties of the generalized hypergeometric function ${}_pF_{q}$

Proof:

The Pocchhammer symbol (rising factorial) is defined as the product: $$(x)_{n} = x(x+1)\cdots(x+n-1) = \prod_{j=0}^{n-1}(x+j)$$ while the generalized hypergeometric function is defined $${}_pF_{q} \left[{a_{1},a_2,...,a_{p}\atop b_{1},b_{2},...,b_{q}}; z \right] = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}(a_{2})_{n}\cdots (a_{p})_n}{(b_{1})_{n}(b_{2})_{n}\cdots (b_{q})_{n} } \frac{z^n}{n!}$$

Note that $(1)_{n} = n!$

The rising factorial obey the duplication formula: $$(x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_{n} \left(\frac{1+x}{2}\right)_{n} \tag{1}$$ and the recursion formula for the argument $$(n+x)(x)_{n} = x(x+1)_{n} \tag{2}$$ Hence $$\begin{align*} S = \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(2n)!^5}{ 2^{10n}n!^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(1)_{2n}^5}{ 2^{10n}(1)_{n}^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)2^{10n}\left(\frac{1}{2}\right)_{n}^{5}(1)_{n}^{5}}{ 2^{10n}(1)_{n}^{10}}\quad \textrm{from (1)}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n4\left(n+\frac{1}{4}\right)\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{ (1)_n(1)_n(1)_n(1)_nn!}\\ =& \sum_{n=0}^{\infty} \frac{\left(\frac{5}{4}\right)_{n}\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{\left(\frac{1}{4}\right)_{n} (1)_n(1)_n(1)_n(1)_n} \frac{(-1)^n}{n!} \quad \textrm{from (2)}\\ =& {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right] \end{align*}$$ One of the Whipple's identities states that $${}_{6}F_{5} \left[{a, 1+\frac{1}{2}a , b,c,d,e \atop \frac{1}{2}a, 1+a-b, 1+a-c, 1+a-d, 1+a-e }; -1\right] = \frac{\Gamma(1-a-d)\Gamma(1+a-e)}{\Gamma(1+a)\Gamma(1+a-d-e)} {}_{3}F_{2}\left[ {1+a-b-c,d,e \atop 1+a-b,1+a-c}; 1\right]$$ Hence $$S = {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right]= \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right]$$ ${}_{3}F_{2}$ obey the Dixon's well poised sum: $${{}_{3}F_{2}}\left[{a,b,c\atop a-b+1,a-c+1};1\right]=\frac{\Gamma\left(\frac{1% }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left(% \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1% \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)},$$ Hence, using $\displaystyle \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$ and $\displaystyle \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ $$\begin{align*} S = \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right] =& \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)} \\ =& \frac{4\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \\ =& \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left(\Gamma(z+1) = z\Gamma(z) \right) \\ =& \frac{2}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left( \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin\left(\pi z\right)}\right)\\ =& \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)} \end{align*}$$ $$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}}$$

Integral of the day XXXIV

Integral involving the golden mean $\phi$

Today we show the proof of the following integral posted by @integralsbot $$\int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15}$$ The proof relies on a "discrete" Laplace transform a technique that we have used in the past here and here

Proof:

Consider the following discrete Laplace transform (proof in Appendix 1): $$1+ 2\sum_{k=1}^{\infty} e^{-kt}\cos(kx) = \frac{\sinh t}{\cosh t -\cos x } \quad t>0$$ If $\cosh t = \frac{\sqrt{5}}{2}$ then $$t = \ln\left(\frac{1+\sqrt{5}}{2} \right) = \ln(\phi)$$ $$\sinh\left(\frac{1+\sqrt{5}}{2} \right) = \frac{1}{2}$$ therefore $$\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x} = 1+ 2\sum_{k=1}^{\infty} \phi^{-k}\cos(kx)$$ where $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$

Hence $$\begin{align*} I = \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx =& \frac{1}{2} \int_0^\pi \frac{x^2}{\frac{\sqrt{5}}{2}-\cos x} dx \\ =& \int_0^\pi \left(\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x}\right) x^2 dx\\ =& \int_0^\pi \left(1+ 2\sum_{k=1}^{\infty} e^{-k\ln(\phi)}\cos(kx)\right) x^2 dx\\ =& \int_{0}^{\pi }\left[ x^2 + 2\sum_{k=1}^{\infty} \phi^{-k} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \phi^{-k} \int _{0}^{\pi} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3} \int _{0}^{k\pi} \cos(w)w^2\right] dx \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ w^2\sin(w)\Big|_{0}^{k\pi} -2\int_{0}^{k\pi} \sin(w)w dw \right] \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(w)w\Big|_{0}^{k\pi} +2 \int_{0}^{\pi k} \cos(w) dw \right] \\ = &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(k \pi )k\pi -2\sin(\pi k)\right] \\ =& \frac{\pi^3}{3} + \underbrace{2\sum_{k=1}^{\infty} \phi^{-k}\frac{(k^2\pi^2-2)\sin(k\pi)}{k^3}}_{\sin(\pi k) = 0 } + \underbrace{4\sum_{k=1}^{\infty} \phi^{-k}\pi \frac{\cos(k \pi)}{k^2} }_{\cos(\pi k) =(-1)^k}\\ =& \frac{\pi^3}{3} + 4\pi \sum_{k=1}^{\infty} \frac{(-\phi^{-1})^k}{k^2} \\ =& \frac{\pi^3}{3} + 4\pi \operatorname{Li}_{2}(-\phi^{-1}) \\ \end{align*}$$ where we used the series expansion for the dilogarithm function $\operatorname{Li}_{2}(z)$: $$\operatorname{Li}_{2}(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2}$$ Landen and Euler proved (proof in the Appendix 2): $$\operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi)$$ Therefore $$\boxed{ \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15} }$$

Appendix 1

$$1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 }$$ Proof: $$\begin{align*} 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{-kt}e^{ikx} \right)\\ =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{k(ix-t)} \right)\\ =& 1+2\Re\left(\frac{1}{e^{t-ix}-1} \right)\\ =& 1+2\Re\left(\frac{e^{-t}}{e^{-ix}-e^{-t}} \right)\\ =& 1+2e^{-t}\Re\left(\frac{1}{\cos x-i\sin x- e^{-t}}\right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{(\cos x-e^{-t})^2 + \sin^2 x} \right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{e^{-2t}-2e^{t}\cos x + 1} \right)\\ =& 1+\Re\left(\frac{\cos x -e^{-t}+ i\sin x}{\cosh t-\cos x} \right)\\ =& \frac{\cosh t- \cos x}{\cosh t- \cos x}+\frac{\cos x-e^{-t}}{(\cosh t- \cos x )} \\ =& \frac{\sinh t}{\cosh t - \cos x} \end{align*}$$

Appendix 2

$$\operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi)$$

Proof:

Consider the following derivative: $$\frac{d}{dx} \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = = -\left[ \ln\left(1+ \frac{x}{1-x}\right)\right]\left(\frac{1}{x} + \frac{1}{1-x} \right)= \ln(1-x)\left(\frac{1}{x} + \frac{1}{1-x} \right)$$ Integrating $$\operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = -\operatorname{Li}_{2}(x) - \frac{1}{2} \ln^2(1-x)$$ $$\Longrightarrow \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \operatorname{Li}_{2}(x) = - \frac{1}{2} \ln^2(1-x) \tag{1}$$ If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ $$\Longrightarrow \operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2} \right) + \operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = - \frac{1}{2} \ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{2}$$ Now we will show that $$\frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3}$$ From the factorization $$(1-x^r) = (1-x)(1-wx)(1-w^2x)\cdots(1-w^{r-1}x)$$ where $\displaystyle w = e^{\frac{i2\pi}{r}}$ Taking logarithms and dividing by $x$ both sides $$\frac{\ln(1-x^r)}{x} = \frac{\ln(1-x)}{x} + \frac{\ln(1-wx)}{x} + \cdots +\frac{\ln(1-w^{r-1}x)}{x}$$ Integrating $$\frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x) + C$$ If $x \to 0$ then $C\to 0$ $$\frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x)$$ If we put $r=2$ $$\frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3}$$ From (1) and (3) we have $$\operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \frac{1}{2}\operatorname{Li}_{2}(x) - \operatorname{Li}_2(-x) = -\frac{1}{2}\ln^2(1-x) \tag{4}$$ If we solve the equation $$\frac{-x}{1-x} = x^2 \Longrightarrow x^2-x+1=0 \Longrightarrow x = \frac{1-\sqrt{5}}{2}$$ Hence from (4) we have $$\frac{3}{2}\operatorname{Li}_{2} \left(\frac{3-\sqrt{5}}{2}\right)- \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = -\frac{1}{2}\ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{5}$$ If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ in the relation (Proof in the Appendix 2 here) $$\operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(1-x) = \frac{\pi^2}{6} -\ln(x)\ln(1-x) \tag{6}$$ we have $$\operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) + \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} -\ln\left(\frac{3-\sqrt{5}}{2}\right)\ln\left(\frac{\sqrt{5}-1}{2}\right) \tag{7}$$ Solving the system given by (5) and (7) we have $$\operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = \frac{\pi^2}{15} - \frac{1}{4}\ln^2\left(\frac{3-\sqrt{5}}{2}\right) \tag{8}$$ Then from (2) and (8) we get $$\operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\frac{\sqrt{5}-1}{2}\right)$$ which is basically the same as $$\operatorname{Li}_{2}\left(-\phi^{-1}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\phi\right)$$

Integral of the day XXXIII

An integral representation of the dilogarithm function

Today we show the proof of this integral posted by @Ali39342137 $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)$$ In general we found $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1$$ In the proof we make use of Fourier series and some properties of the dilogarithm function $\operatorname{Li}_{2}(z)$.

Proof:

Consider the following Fourier series (Proof in the Appendix 1): $$\sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1$$ Then, if $p = -q$ $$\sum_{k=1}^{\infty } \frac{(-1)^kq^{k}\sin(kx)}{k} = \arctan\left(\frac{-q\sin x}{1+q\cos x}\right) = - \arctan\left(\frac{q\sin x}{1+q\cos x}\right)$$ Therefore $$\arctan\left(\frac{\sin x}{\frac{1}{q}+\cos x}\right) = \sum_{k=1}^{\infty } \frac{(-1)^{k+1}q^{k}\sin(kx)}{k} \quad \frac{1}{q}\geq 1$$ Hence $$\begin{align*} I = \int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx =& \int_{0}^{\pi} x \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(kx)}{k}dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k} \int_{0}^{\pi} x\sin(kx)dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \int_{0}^{k\pi} w\sin(w)dx \quad (w \mapsto kx)\\ \stackrel{IBP}{=}& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-\cos(x)x\Big|_{0}^{k\pi}+ \int_{0}^{k\pi}\cos(w)dx \right] \\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-k\pi\cos(\pi k) + \sin(\pi k) \right] \\ =& \pi \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k}\left(\frac{1}{a}\right)^{k}\cos(k\pi) }{k^2}}_{\cos(k \pi) = (-1)^k} + \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(\pi k) }{k^3}}_{\sin(k\pi)=0} \\ =& \pi \sum_{k=1}^{\infty } \frac{\left(\frac{1}{a}\right)^{k} }{k^2} \\ =& \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \end{align*}$$ If $a = 2$ $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{2}\right)$$ From the relation (proof in the Appendix 2) $$\operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z)$$ if we put $\displaystyle z = \frac{1}{2}$ $$\operatorname{Li}_{2}\left(\frac{1}{2} \right) = \frac{\pi^2}{12} -\frac{1}{2}\ln^2\left(2\right)$$ Therefore $$\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)}$$ $$\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1 }$$

Appendix 1 $$\sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1$$

Proof

Take the principal branch of the $\ln(z)$ function. Hence $$\begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& \Im\left( \sum_{k=1}^{\infty } \frac{(pe^{ix})^k}{k} \right) = \Im\left( -\ln(1-pe^{ix})\right)\\ =& -\arg\left(1-pe^{ix}\right)\\ =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ \end{align*}$$ Since $|p|\lt 1$ then $1-p\cos(x)>0$ therefore $$\begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ =& -\arctan\left( \frac{-p\sin x}{1-p\cos(x)}\right) \\ =& \arctan\left( \frac{p\sin x}{1-p\cos(x)}\right) \quad \left(\arctan \textrm{ is odd } \right) \end{align*}$$

Appendix 2 $$\operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z)$$

Proof $$\begin{align*} \operatorname{Li}_{2}(z) = &\int_{z}^{0} \frac{\ln(1-t)}{t} dt \stackrel{IBP}{=} -\ln(z)\ln(1-z) -\int_{0}^{z}\frac{\ln (t)}{1-t}dt\\ =& -\ln(z)\ln(1-z) + \int_{1}^{1-z}\frac{\ln (1-w)}{w}dw \quad \left( w\mapsto 1-t \right)\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw + \int_{0}^{1-z}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw - \int_{1-z}^{0}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \operatorname{Li}_{2}(1) - \operatorname{Li}_{2}(1-z)\\ \end{align*}$$ Therefore $$\operatorname{Li}_{2}(z) + \operatorname{Li}_{2}(1-z) = -\ln(z)\ln(1-z) + \zeta(2) = -\ln(z)\ln(1-z) + \frac{\pi^2}{6}$$

Application of Ramanujan's master theorem II

A useful Mellin transform related to the Bernoulli and zeta numbers

Today we show the proof of this Mellin transform posted by @integralsbot $$\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =\Gamma(s-n)\zeta(s-n) \quad 0\lt \Re(s)\lt1$$ The proof relies on the Ramanujan's master theorem, some properties of the Bernoulli numbers and the binomial coefficient.

We also show the proof of this nice application: an integral representation for the Apery's constant (also posted by @integralsbot ) : $$\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2}$$

Proof:

We will apply the Ramanujan master theorem: If a complex-valued function $f(x)$ has an expansion of the form $$f(x)=\sum_{k=0}^\infty \frac{\,\varphi(k)\,}{k!}(-x)^k$$ then the Mellin transform of $f(x)$ is given by $$\int_0^\infty x^{s-1} f(x) dx = \Gamma(s)\,\varphi(-s)$$ where $\Gamma(s)$ is the gamma function.

The Bernoulli numbers are defined through the generating function $$\frac{x}{e^x -1} = \sum_{k=0}^{\infty} \frac{ B_{k}x^k}{k!}$$ Therefore $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} = & \sum_{k=0}^{\infty} \frac{B_{k} x^k}{k!} - \sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\\ =& \sum_{k=n+1}^{\infty} \frac{B_{k} x^k}{k!}\\ =& \sum_{j=0}^{\infty} \frac{B_{j+n+1}x^{j+n+1}}{(j+n+1)!} \end{align*}$$ The zeta function are related to the Bernoulli numbers through the following expression $$B_n = -\zeta(1-n)n \quad n=2,3,4,..$$ Hence $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} = &\sum_{j=0}^{\infty} \frac{B_{j+n+1}x^{j+n+1}}{(j+n+1)!} \\ =& \sum_{j=0}^{\infty} \frac{\zeta(-j-n)(j+n+1)x^{j+n+1}}{(j+n+1)!}\\ =& \sum_{j=0}^{\infty} \frac{\zeta(-j-n)x^{j+n+1}}{(j+n)!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n) j!(-1)^j}{(j+n)!} \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n) j!n!(-1)^j}{n! (j+n)!} \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{n! \binom{n+j}{n} } \frac{(-x)^{j}}{j!} \end{align*}$$ The formula $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \tag{1}$$ relate the binomial coefficient to the Pochhammer polynomial (rising factorial) while the Pochhammer polynomial obey the reflection formula $$(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{2}$$ Hence $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{n! \binom{n+j}{n} } \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{(j+1)_{n}} \frac{(-x)^{j}}{j!} \quad \textrm{from (1)}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!} \quad \textrm{from (2)}\\ \end{align*}$$ Back to the integral: $$\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx = \int_{0}^{\infty} x^{s-1} \left(\sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!}\right) dx$$ Denote $$f(x) = \sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!}$$ $$\varphi(j) =\frac{\zeta(-j-n)}{(-j-n)_{n}}$$ Applying the Ramanujan's master theorem $$\begin{align*} I = \int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =& \Gamma(s)\varphi(-s)\\ =& \Gamma(s)\frac{\zeta(s-n)}{(s-n)_{n}} \end{align*}$$ Finally, recall that the rising facotiral can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Hence $$I = \Gamma(s)\frac{\zeta(s-n)}{(s-n)_{n}} = \Gamma(s)\frac{\zeta(s-n)\Gamma(s-n)}{\Gamma(s)} = \Gamma(s-n)\zeta(s-n)$$ Therefore, we can conclude $$\boxed{\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =\Gamma(s-n)\zeta(s-n) \quad 0<\Re(s)<1 }$$

Application: Integral representation of the Apery's constant

$$\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2}$$

Proof

Put $n=2$ in the Mellin transform, then $$\int_0^{\infty} x^{s-4}\left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)dx = -\Gamma(s-2)\zeta(s-2)$$ Under some technical conditions the Mellin transform converges uniformly and we can take the limit as $s \to 0$ $$\lim_{s \to 0} \int_0^{\infty} x^{s-4}\left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)dx = \int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2)$$ What is the limit on the right hand side? $$\begin{align*} \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2) =& \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2)\\ =& \lim_{t \to -2} -\Gamma(t)\zeta(t)\\ =& \lim_{t \to -2} -\frac{\zeta(t)}{\frac{1}{\Gamma(t)}}\\ =& \lim_{t \to -2} -\frac{\zeta(t)}{\frac{t(t+1)(t+2)}{\Gamma(t+3)}} \quad \left(\Gamma(t+1) = t\Gamma(t) \right)\\ =& -\frac{\lim_{t \to -2} \zeta'(t)}{\lim_{t \to -2} \frac{3t^2+6t+2-(t^2+3t+2)\psi(t+3)}{\Gamma(t+3)}} \quad \textrm{(L'Hôpital's rule )}\\ =& -\frac{\zeta'(-2)}{2}\\ \end{align*}$$ The derivative of the zeta function at the negative even integers is given by $$\zeta^{\prime}(-2n) = (-1)^n \frac {(2n)!} {2 (2\pi)^{2n}} \zeta (2n+1)$$ If we put $n=1$ $$\zeta^{\prime}(-2) = - \frac {\zeta (3)} { 4\pi^2}$$ Hence $$\lim_{s \to 0} -\Gamma(s-2)\zeta(s-2) = \frac {\zeta (3)} { 8\pi^2}$$ Therefore we can conclude $$\boxed{\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2} }$$