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