Integral of the day XXVII

Integral involving the Gamma functon $\Gamma(\cdot)$

Today we show the proof of this integral posted by @Ali39342137 $$\int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx = \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k}\quad n> \frac{k}{2}$$ This is a generalization for this other integral posted by @integralsbot with $n=s$ and $k=1$: $$\int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^s dx = \frac{s}{2s+1}\sqrt{\frac{\pi}{2}} \frac{\Gamma\left(\frac{s}{2}-\frac{1}{4}\right)}{\Gamma\left(\frac{s}{2}+\frac{1}{4}\right)}$$ We previously posted another similar generalization here

Proof:

$$\begin{align*} I = \int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx =& \frac{1}{2}\int_{0}^{\infty} \left(\sqrt{1+w^2}-w\right)^nw^{\frac{k}{2}-1} dw \quad ( w \mapsto x^2) \\ =& \frac{1}{2^{\frac{k}{2}+1}}\int_{0}^{\infty} t^{n-\frac{k}{2}-1}(1-t^2)^{\frac{k}{2}-1}(t^2+1) \quad (t \mapsto \sqrt{1+w^2}-w ) \\ =& \frac{1}{2^{\frac{k}{2}+1}}\left[\int_{0}^{\infty} t^{n-\frac{k}{2}+1}(1-t^2)^{\frac{k}{2}-1} dt +\int_{0}^{\infty} t^{n-\frac{k}{2}-1}(1-t^2)^{\frac{k}{2}-1} dt\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\int_{0}^{\infty} s^{\frac{n}{2}-\frac{k}{4}}(1-s)^{\frac{k}{2}-1} ds + \int_{0}^{\infty} s^{\frac{n}{2}-\frac{k}{4}-1}(1-s)^{\frac{k}{2}-1} ds\right] \quad ( s \mapsto t^2)\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[B\left(\frac{n}{2}-\frac{k}{4}+1, \frac{k}{2}\right)+ B\left(\frac{n}{2}-\frac{k}{4}, \frac{k}{2}\right)\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}+1\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}+1\right)} + \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\right]\\ =& \frac{1}{2^{\frac{k}{2}+2}}\left[\frac{\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\left(\frac{n}{2}+\frac{k}{4}\right)\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)} + \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\right]\\ =& \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}+2}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\left[\frac{\left(\frac{n}{2}-\frac{k}{4}\right)}{\left(\frac{n}{2}+\frac{k}{4}\right)} + 1\right]\\ =& \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k} \end{align*}$$ $$\boxed{ \int_{0}^{\infty} \left(\sqrt{1+x^4}-x^2\right)^nx^{k-1} dx = \frac{\Gamma\left(\frac{n}{2}-\frac{k}{4}\right)\Gamma\left(\frac{k}{2}\right)}{2^{\frac{k}{2}}\Gamma\left(\frac{n}{2}+\frac{k}{4}\right)}\frac{n}{2n+k}\quad n> \frac{k}{2} }$$

Golden ratio II

Nice hypergeometric series involving the Fibonacci numbers $F_{2k+3}$

Today we show the proof of this nice series proposed by @diegorattaggi $$\sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = 2^k F_{2k+3}$$ For the proof we will use some properties of the rising factorial (Pochhammer polynomial), the Gaussian hypergeometric function and the Binet's formula for Fibonacci numbers.

Proof:

$$S = \sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = \sum_{n=1}^{\infty} \frac{(n+k-1)!}{(n-1)!k!} \frac{F_{n}}{2^n} = \sum_{n=1}^{\infty} \frac{\Gamma(n+k)}{(n-1)!k!} \frac{F_{n}}{2^n}$$ Recall that the pochhammer polynomial (rising factorial) can be expressed as the quotient of two gamma functions: $$\frac{\Gamma(n+x)}{\Gamma(x)} = (x)_{n}$$ where $$(x)_{n} = x(x+1)(x+2)\cdots (x+n-1)$$ Hence $$S = \sum_{n=1}^{\infty} \frac{\Gamma(n+k)}{(n-1)!k!} \frac{F_{n}}{2^n} = \frac{\Gamma(k)}{k!}\sum_{n=1}^{\infty} \frac{(k)_{n}}{(n-1)!} \frac{F_{n}}{2^n} = \frac{1}{k}\sum_{n=1}^{\infty} \frac{(k)_{n}}{(n-1)!} \frac{F_{n}}{2^n}$$ If we let the change of variable $n=j+1$ $$S = \frac{1}{2k}\sum_{j=0}^{\infty} \frac{(k)_{j+1}}{j!} \frac{F_{j+1}}{2^{j}}$$ From the recursion formula for the degree of the Pochhammer polynomial: $$(x)_{n+1} = x(x+1)_{n}$$ we have $$S = \frac{1}{2k}\sum_{j=0}^{\infty} \frac{(k)_{j+1}}{j!} \frac{F_{j+1}}{2^{j}} = \frac{1}{2}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{F_{j+1}}{2^{j}}$$ From the Binet's formula $$F_{j+1} = \frac{\phi^{j+1}-\psi^{j+1}}{\sqrt{5}}$$ $$S = \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\phi^{j}}{2^{j}} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\psi^{j}}{2^{j}}$$ Multiplying and dividing by $(1)_{j}$ we have an hypergeometric representation: $$\begin{align*} S =& \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\phi^{j}}{2^{j}} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}}{j!} \frac{\psi^{j}}{2^{j}} \\ =& \frac{\phi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}(1)_{j}}{(1)_{j}} \frac{\left(\frac{\phi}{2}\right)^j}{j!} - \frac{\psi}{2\sqrt{5}}\sum_{j=0}^{\infty} \frac{(k+1)_{j}(1)_{j}}{(1)_{j}} \frac{\left(\frac{\psi}{2} \right)^j}{j!}\\ =& \frac{\phi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\phi}{2}\right)- \frac{\psi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\psi}{2}\right) \end{align*}$$ This is a special case of the Gaussian hypergeometric fucntion (generalized Binomial theorem): $$F(a,b;a;z ) = (1-z)^{-b}$$ If we put $a=1, b=k+1$ $$S = \frac{\phi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\phi}{2}\right)- \frac{\psi}{2\sqrt{5}}F\left(1,k+1;1;\frac{\psi}{2}\right) = \frac{\phi}{2\sqrt{5}}\left(1-\frac{\phi}{2} \right)^{-k-1}-\frac{\psi}{2\sqrt{5}}\left(1-\frac{\psi}{2} \right)^{-k-1}$$ Note $$\begin{align*} \frac{\phi}{2\sqrt{5}}\left(1-\frac{\phi}{2} \right)^{-k-1} =& \frac{\phi}{2\sqrt{5}}\left(\frac{2}{2-\phi} \right)^{k+1}\\ =& \frac{\phi}{2\sqrt{5}}\left(3+\sqrt{5}\right)^{k+1}\\ =& \frac{\phi}{2\sqrt{5}}\left(2+2\phi\right)^{k+1}\\ =& \frac{\phi 2^{k+1}}{2\sqrt{5}}\left(1+\phi\right)^{k+1}\\ =& \frac{\phi 2^{k+1}}{2\sqrt{5}}(\phi^2)^{k+1} \quad (1+\phi = \phi^2)\\ =& \frac{2^k \phi^{2k+3}}{\sqrt{5}} \end{align*}$$ $$\begin{align*} \frac{\psi}{2\sqrt{5}}\left(1-\frac{\psi}{2} \right)^{-k-1} = & \frac{\psi}{2\sqrt{5}}\left(1-\frac{1-\phi}{2} \right)^{-k-1} \quad (\psi = 1-\phi)\\ =& \frac{\psi}{2\sqrt{5}}\left(\frac{2}{1+\phi} \right)^{k+1}\\ =& \frac{2^k\psi}{\sqrt{5}}\left(\frac{1}{\phi^2} \right)^{k+1} \quad (\phi+1 = \phi^2) \\ =& \frac{2^k\psi}{\sqrt{5}}\left(\psi^2\right)^{k+1}\\ =& \frac{2^k\psi^{2k+3}}{\sqrt{5}} \end{align*}$$ Hence $$S = \frac{2^k \phi^{2k+3}}{\sqrt{5}} -\frac{2^k}{\sqrt{5}}\psi^{2k+3} = 2^{k}\left[\frac{\phi^{2k+3}}{\sqrt{5}} -\frac{\psi^{2k+3}}{\sqrt{5}}\right] = 2^kF_{2k+3}$$ Therefore $$\boxed{ \sum_{n=1}^{\infty} \binom{n+k-1}{k} \frac{F_{n}}{2^n} = 2^k F_{2k+3}}$$

Integral of the day XXVI

Integral involving the Dirichlet eta function $\eta(v)$

Today we show the proof of this integral proposed by @Ali39342137: $$\int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)$$ For the proof we use the expansion in partial fractions of the hyperbolic cosecant function and the derivative of the Dirichlet eta function.

Proof:

Recall that the hyperbolic cosecant can be expanded as partial fractions: $$\operatorname{csch} x = \frac{1}{\sinh x} = \frac{1}{x}-\frac{2x}{\pi^2+x^2}+ \frac{2x}{4\pi^2+x^2}-\frac{2x}{9\pi^2+x^2} + \cdots = \frac{1}{x} + 2\sum_{j=1}^{\infty} \frac{(-1)^jx}{j^2\pi^2+x^2}$$ Hence, $$\begin{align*} \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx =& \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{x} - 2\sum_{j=1}^{\infty} \frac{(-1)^jx}{j^2\pi^2+x^2}\right)\frac{\ln(x)}{x} dx \\ =& -2\int_{0}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^j\ln(x)}{j^2\pi^2+x^2} dx \\ =& -2\sum_{j=1}^{\infty}\int_{0}^{\infty} (-1)^j\frac{\ln(x)}{j^2\pi^2+x^2} dx \\ =& -\frac{2}{\pi^2}\sum_{j=1}^{\infty}\frac{(-1)^j}{j^2} \int_{0}^{\infty} \frac{\ln(x)}{1+\left(\frac{x}{j\pi}\right)^2} dx \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \int_{0}^{\infty} \frac{\ln(j\pi w)}{1+w^2} dx \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \int_{0}^{\infty} \frac{\ln(j\pi)+\ln(w) }{1+w^2} dw \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\ln(j\pi)\int_{0}^{\infty} \frac{1}{1+w^2} dw + \int_{0}^{\infty} \frac{\ln(w) }{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \int_{0}^{\infty} \frac{\ln(w) }{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \int_{0}^{\infty} \frac{\frac{d}{dt}\Big|_{t=0+} w^t}{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} \frac{ w^t}{1+w^2} dw\right] \\ \end{align*}$$ Recall the integral representation of the secant function: $$\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2x}{\pi}}}{y^2+1} dy \quad \left|x\right|\lt \frac{\pi}{2}$$ If we put $\displaystyle x = \frac{t\pi}{2}$ $$\sec\left( \frac{t\pi}{2}\right) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{t}}{y^2+1} dy$$ Hence $$\int_{0}^{\infty} \frac{y^{t}}{y^2+1} dy = \frac{\pi}{2}\sec\left( \frac{t\pi}{2}\right)$$ Therefore $$\begin{align*} I =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} \frac{ w^t}{1+w^2} dw\right]\\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+} \frac{\pi}{2}\sec\left( \frac{t\pi}{2}\right)\right]\\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \underbrace{\lim_{t \to 0+} \frac{\pi^2}{4}\sec\left( \frac{t\pi}{2}\right)\tan\left( \frac{t\pi}{2}\right)}_{=0}\right]\\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j\pi)}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\left[\ln(j)+ \ln(\pi)\right]}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\ln(2) \\ \end{align*}$$ Recall the definition of the Dirichlet eta funcion: $$\eta(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j^v}$$ Taking the derivative $$\eta'(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j^v}$$ Therefore $$\lim_{v \to 1} \eta'(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j}$$ Now recall the relationship: $$\eta(v) = (1-2^{1-V}) \zeta(v)$$ Hence $$\eta'(v) = 2^{-v}\left[(2^v-2)\zeta'(v) + \ln(4)\zeta(v) \right]$$ Taking the limit as $v \to 1$ can be proven that: $$\lim_{v \to 1} \eta'(v) = \gamma\ln(2) -\frac{1}{2}\ln^2(2)$$ Therefore $$\sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j} = \gamma\ln(2) -\frac{1}{2}\ln^2(2)$$ Then we have $$I = \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\ln(2) = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)$$ Therefore we can conclude $$\boxed{ \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)}$$

Integral of the day XXV

Integral related to the reciprocal beta function

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = 2\pi\ln(2)$$ The proof will rely on the integral representation of the reciprocal beta function.

Proof:

Recall the following integral representation of the reciprocal beta function (Proof in the Appendix): $$\int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v,a\in \mathbb{R} , \; \; v>0$$ $$\int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v,a\in \mathbb{R} , \; \; v>0$$ If we put $\displaystyle v = \frac{1}{2}$ $$\int_{0}^{\pi} \frac{ \sin(ax)}{\sqrt{\sin x} } dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{\frac{1}{2}-1}vB\left(\frac{1+a}{2}+\frac{1}{4},\frac{1-a}{2}+\frac{1}{4}\right)}$$ $$\int_{0}^{\pi} \frac{ \cos(ax)}{\sqrt{\sin x} } dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{\frac{1}{2}-1}vB\left(\frac{1+a}{2}+\frac{1}{4},\frac{1-a}{2}+\frac{1}{4}\right)}$$ Differentiating with respect to $a$: $$\frac{d}{da} \int_{0}^{\pi} \frac{ \sin(ax)}{\sqrt{\sin x} } dx = \int_{0}^{\pi} \frac{ x\cos(ax)}{\sqrt{\sin x} } dx = \frac{\sqrt{2}\pi^2\cos\left(\frac{\pi a}{2}\right) + \sqrt{2}\pi \sin\left(\frac{\pi a}{2}\right)\psi\left(\frac{3}{4}-\frac{a}{2}\right)-\sqrt{2}\pi \sin\left(\frac{\pi a}{2}\right)\psi\left(\frac{a}{2}+\frac{3}{4}\right)}{B\left(\frac{a}{2}+\frac{3}{4}, \frac{3}{4}-\frac{a}{2}\right)}$$ $$\frac{d}{da} \int_{0}^{\pi} \frac{ \cos(ax)}{\sqrt{\sin x} } dx = -\int_{0}^{\pi} \frac{ x\sin(ax)}{\sqrt{\sin x} } dx = \frac{\sqrt{2}\pi^2\sin\left(\frac{\pi a}{2}\right) - \sqrt{2}\pi \cos\left(\frac{\pi a}{2}\right)\psi\left(\frac{3}{4}-\frac{a}{2}\right)+\sqrt{2}\pi \cos\left(\frac{\pi a}{2}\right)\psi\left(\frac{a}{2}+\frac{3}{4}\right)}{B\left(\frac{a}{2}+\frac{3}{4}, \frac{3}{4}-\frac{a}{2}\right)}$$ Taking the limit as $\displaystyle a \to \frac{1}{2}$ $$\int_{0}^{\pi} \frac{x\cos\left(\frac{x}{2}\right)}{\sqrt{\sin x}} dx = \frac{\pi\gamma+\pi^2+\pi \psi\left(\frac{1}{2}\right)}{2}$$ $$-\int_{0}^{\pi} \frac{ x\sin\left(\frac{x}{2}\right)}{\sqrt{\sin x} } dx = \frac{\pi\gamma-\pi^2+\pi \psi\left(\frac{1}{2}\right)}{2}$$ From the formula $$\psi\left(\frac{1}{2}-J\right) = \psi\left(\frac{1}{2}+J\right) = -\gamma -\ln(4) +\sum_{j=1}^{J} \frac{2}{2j-1} \quad J=0,1,2,...$$ if we put $J=0$ we have $$\psi\left(\frac{1}{2}\right) + \gamma = -\ln(4)$$ Therefore $$\int_{0}^{\pi} \frac{x\cos\left(\frac{x}{2}\right)}{\sqrt{\sin x}} dx = \frac{\pi^2-\pi\ln(4)}{2}$$ $$-\int_{0}^{\pi} \frac{ x\sin\left(\frac{x}{2}\right)}{\sqrt{\sin x} } dx = \frac{-\pi^2-\pi\ln(4)}{2}$$ Hence $$\int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = \frac{\pi^2}{2} + \frac{\pi\ln(4)}{2} -\frac{\pi^2}{2} +\frac{\pi\ln(4)}{2} = 2\pi\ln(2)$$ Therefore we can conclude $$\boxed{\int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = 2\pi\ln(2) }$$

Appendix:

Here is the proof of the following integrals with $a,b \in \mathbb{R}$: $$\int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v>0$$ $$\int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v>0$$

Proof:

Let $a,b \in \mathbb{R}$ $$\begin{align*} I= \int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx =& \int_{0}^{\pi} e^{iax}\left(\frac{e^{ix}+e^{-ix}}{2i} \right)^{v-1}dx \\ =& \frac{1}{(2i)^{v-1}}\int_{0}^{\pi} e^{iax} \sum_{j=0}^{\infty} \binom{v-1}{j}(-1)^j e^{ix(v-1-2j)} dx\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}(-1)^j\int_{0}^{\pi} e^{ix(a+v-1-2j)} dx\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j}{i(a+v-1-2j)}\int_{0}^{i\pi(a+v-1-2j)} e^{w}dw\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j\left(e^{i\pi (a+v-1-2j)}-1\right)}{i(a+v-1-2j)}\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j\left(e^{i\pi (a+v-1)}-1\right)}{i(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j}{(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(v-j)_{j}}{j!}\frac{(-1)^j}{(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(1-v)_{j}}{(a+v-1-2j)j!} \end{align*}$$ Note that $$2j-a-v+1 = \frac{-(a+v-1)\left(\frac{3-a-v}{2}\right)_{j}}{\left(\frac{-a-v+1}{2}\right)_{j}}$$ Hence $$\begin{align*} I =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(1-v)_{j}}{(a+v-1-2j)j!}\\ =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} \sum_{j=0}^{\infty} \frac{(1-v)_{j}\left(\frac{-a-v+1}{2}\right)_{j}}{\left(\frac{3-a-v}{2}\right)_{j}}\\ =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} F\left(1-v,\frac{-a-v+1}{2};\frac{3-a-v}{2};1 \right) \end{align*}$$ From the Gauss's Hypergeometric Theorem: $$F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \quad c>a+b$$ Therefore, supposing $v>0$: $$\begin{align*} I =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} \frac{\Gamma\left(\frac{3-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v}i^v}\frac{1}{-\frac{(1-a-v)}{2}} \frac{\left(\frac{1-a-v}{2}\right)\Gamma\left(-\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&-\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v}i^v} \frac{\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ \end{align*}$$ Note $$\Im (e^{i\pi (a+v-1)}-1)i^{-v} = -2\sin\left(\frac{\pi a}{2} \right)\cos\left(\frac{\pi(a+v)}{2}\right)$$ $$\Re (e^{i\pi (a+v-1)}-1)i^{-v} = -2\cos\left(\frac{\pi a}{2} \right)\sin\left(\frac{\pi(a+v)}{2}\right)$$ and recall the Euler's reflection formula for $\cos(\pi t)$: $$\Gamma\left(\frac{1}{2}+t\right)\Gamma\left(\frac{1}{2}-t\right) = \frac{\pi}{\cos(\pi t)}$$ Therefore $$\begin{align*} \int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \Im \left(\int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx \right) = & \frac{ \sin\left(\frac{\pi a}{2} \right)\cos\left(\frac{\pi(a+v)}{2}\right)}{2^{v-1}} \frac{\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{ \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}} \frac{\pi\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1+a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)} \quad \textrm{(Euler's reflection formula)}\\ =&\frac{ \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}} \frac{\pi \Gamma(v)}{\Gamma\left(\frac{1+a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \end{align*}$$ In a very similar way we can show : $$\int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \Re \left(\int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx \right) = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)}$$

Generalized hypergeometric functions VIII

Nice integral related to the Gauss hypergeometric function $F(a,b;c;z)$

Today we show the proof of this nice integral without contour integration as proposed by @AlbahariRicardo. $$\int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \quad a,b\in \mathbb{C}, \quad \Re(b)>-1$$ The proof relies on the Gauss theorem for the function that have his name $F(a,b;c;z)$, the proof of this theorem neither relies on contour integration.

Update: We updated this post to cover the case $a,b\in \mathbb{C}$.

Proof:

Let $a,b \in \mathbb{C}$ $$\begin{align*} I= \int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = &\frac{1}{2} \int_{0}^{\frac{\pi}{2}} (e^{ia\theta}+e^{-ia\theta})\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta - \frac{1}{2} \underbrace{\int_{0}^{-\frac{\pi}{2}} e^{ia\vartheta}\left( \frac{e^{i\vartheta }+e^{-i\vartheta}}{2}\right)^b d\vartheta}_{\vartheta \mapsto -\theta}\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta + \frac{1}{2} \int_{-\frac{\pi}{2}}^{0} e^{ia\vartheta}\left( \frac{e^{i\vartheta }+e^{-i\vartheta}}{2}\right)^b d\vartheta\\ =& \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta \\ =& \frac{1}{2^{b+1}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{ia\theta}\sum_{j=0}^{\infty}\binom{b}{j} e^{i\theta(b-2j)} d\theta \quad \textrm{(Generalized binomial theorem)} \\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{i\theta(a+b-2j)} d\theta \\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{1}{i(a+b-2j)} \int_{-\frac{i\pi(a+b-2j)}{2}}^{\frac{i\pi(a+b-2j)}{2}} e^{w} dw \quad (w \mapsto i\theta(a+b-2j), \quad a+b\neq 2j \quad j\in \mathbb{N})\\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{e^{\frac{i\pi(a+b-2j)}{2}}-e^{\frac{-i\pi(a+b-2j)}{2}}}{i(a+b-2j)}\\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{e^{\frac{i\pi(a+b)}{2}}e^{-i\pi j} -e^{\frac{-i\pi(a+b)}{2}}e^{i\pi j}}{i(a+b-2j)}\\ =& \frac{1}{2^{b}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{(-1)^j}{(a+b-2j)} \frac{e^{\frac{i\pi(a+b)}{2}} -e^{\frac{-i\pi(a+b)}{2}}}{2i}\\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{(-1)^j}{(a+b-2j)} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty}\frac{(b-j+1)_{j}}{j!} \frac{(-1)^j}{(a+b-2j)} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty} \frac{(-b)_{j}}{(a+b-2j)j!} \\ \end{align*}$$ Note that $$2j-a-b = \frac{-(a+b)\left(\frac{2-a-b}{2}\right)_{j}}{\left(\frac{-a-b}{2}\right)_{j}}$$ Hence $$\begin{align*} I = \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty} \frac{(-b)_{j}}{(a+b-2j)j!} = & \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}(a+b)}\sum_{j=0}^{\infty} \frac{(-b)_{j}\left(\frac{-a-b}{2}\right)_{j}}{\left(\frac{2-a-b}{2}\right)_{j}}\frac{1}{j!} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}(a+b)}F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right) \\ \end{align*}$$ where $F(a,b;c;z)$ is the Gauss hypergeometric function which satisfies th Gauss theorem: $$F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \quad \Re(c-a+b)>0$$ Applying the theorem to $\displaystyle F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right)$ and supposing $\Re(b)>-1$ $$F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right) = \frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)\Gamma(1)}$$ Hence $$\begin{align*} I =&\frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^b(a+b)}\frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)}\\ =& \frac{1}{2^b(a+b)}\left[\frac{\pi}{\Gamma\left(\frac{2-a+b}{2}\right) \Gamma\left(\frac{a+b}{2}\right)}\right]\frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)} \quad \textrm{(Euler reflection formula)} \\ =& \frac{\pi\Gamma(b+1)}{2^b(a+b)\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{a+b}{2}\right)}\\ =& \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \end{align*}$$ The cases $b=a=2n$, $2m=a>b=2n$ and $2m=a\lt b=2n$ can easily be obtained as limiting cases of $$\int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right) } \quad \Re(b)>-1, a+b\neq 2j \quad j\in \mathbb{N}$$ Hence, we can conclude $$\boxed{\int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \quad a,b\in \mathbb{C}, \Re(b)>-1}$$

Integral of the day XXIV

A miserious integral related to Lobachevsky

Today we show the proof of this strange integral posted by @integralsbot $$\int_{0}^\infty \frac{\sin(\tan(x))}{x} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right)$$ The trick to solve this integral is a variation to the Lobachevsky integral formula for odd functions and the application of the residue theorem.

Proof:

Recall the Lobachevsky integral formula for odd functions:

Theorem. Let $f(x)$ an odd, complex-valued, $\pi$-periodic function such that $\displaystyle \frac{f(x)}{\tan x}$ is absolutely integrable over a single period.

Therefore $$\int_{0}^{\infty} \frac{f(x) }{x}dx = \int_{0}^{\frac{\pi}{2}} \frac{f(x)}{\tan\left(x \right)} dx \tag{*}$$ Back to the integral:

Note that the function $$f(x) = \sin(\tan(x))$$ is an odd function that satisfies all the hypothesis of the theorem.

Hence $$\begin{align*} I =& \int_{0}^\infty \frac{\sin(\tan(x))}{x} dx \\ = & \int_{0}^{\frac{\pi}{2}} \frac{\sin(\tan(x))}{\tan(x)} dx \quad \textrm{ from (*) } \\ =& \int_{0}^{\frac{\pi}{2}} \frac{\sin(w)}{w(1+w^2)} dw \quad \left(w \mapsto \tan x\right)\\ =& \int_{0}^\infty \frac{\sin (w)}{w} dw -\int_{0}^{\infty} \frac{w\sin(w)}{1+w^2} dw \\ =& \int_{0}^\infty \frac{\sin (w)}{w} dw -\frac{1}{2}\int_{-\infty}^{\infty} \frac{w\sin(w)}{1+w^2} dw \quad (\textrm{the integrand is even}) \\ =& \underbrace{\int_{0}^\infty \frac{\sin (w)}{w} dw}_{\textrm{Dirichlet integral}} - \Im \left(\underbrace{\int_{-\infty}^{\infty} \frac{we^{iw}}{1+w^2} dw }_{J}\right) \\ =& \frac{\pi}{2}- \Im \left(\underbrace{\int_{-\infty}^{\infty} \frac{we^{iw}}{1+w^2} dw }_{J}\right) \\ \end{align*}$$ The first integral is the classic Dirichlet integral, the second is a little bit more difficult: Let $R>1$ and consider the following complex function $$f(z)= \frac{ze^{iz}}{1+z^2}$$ This function has two poles at $z= \pm i$.

Now consider the following contour (half upper circle): $$C_{1} = t \quad t\in (-R,R)$$ $$C_{2} = Re^{i\theta} \quad \theta \in \left[0,\pi\right]$$ Hence $$\oint_{C=C_{1}\cup C_{2}} \frac{ze^{iz}}{1+z^2} dz = \int_{C_{1}} f(z) dz + \oint_{C_{2}} f(dz)dz = \int_{-R}^{R} \frac{te^{it}}{1+t^2} dt + \int_{0}^{\pi} \frac{iR^2e^{2i\theta}e^{iRe^{i\theta}}}{1+Re^{i\theta}} d\theta$$ We will show that the integral in the right hand side will vanish as $R\to \infty$. For that end we will make use of the Jordan's Lemma:

Jordan's lemma. Consider a complex-valued, continuous function $f$, defined on a semicircular contour $$C_R = \{R e^{i \theta} \mid \theta \in [0, \pi]\}$$ of positive radius $R$ lying in the upper half-plane, centered at the origin. If the function $f$ is of the form $$f(z) = e^{i a z} g(z) , \quad z \in C_R ,$$ with a positive parameter $a$, then Jordan's lemma states the following upper bound for the contour integral: $$\left| \int_{C_R} f(z) \, dz \right| \le \frac{\pi}{a} M_R \quad \text{where} \quad M_R := \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right|$$ with equality when $g$ vanishes everywhere, in which case both sides are identically zero.

Applying the lemma to $f(z)$: $$f(z)= \frac{ze^{iz}}{1+z^2} = e^{iz} g(z) \quad \textrm{ where } g(z) = \frac{z}{1+z^2}$$ Hence $$\left| \int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz \right| \le \pi M_{R}$$ where $$\begin{align*} M_R = \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right| = &\max_{\theta \in [0,\pi]} \left| \frac{Re^{i\theta}}{1+R^2e^{i2\theta}} \right|\\ =& \max_{\theta \in [0,\pi]} \frac{R}{\left| 1+R^2e^{i2\theta}\right|} \\ =& \max_{\theta \in [0,\pi]} \frac{R}{\left| 1-(-R^2e^{i2\theta})\right|} \\ \leq & \frac{R}{\left| 1-R^2\right|}\\ =& \frac{R}{R^2-1} \quad (R>1) \end{align*}$$ Therefore $$\left| \int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz \right| \le \pi M_{R} \leq \frac{\pi R}{R^2-1}$$ Taking the limit as $R \to \infty$ $$\int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz =\int_{0}^{\pi} \frac{iR^2e^{2i\theta}e^{iRe^{i\theta}}}{1+Re^{i\theta}} d\theta= 0$$ Hence, since the only pole in the upper half circle is $z=i$ we have by the residue theorem $$J = \int_{-\infty}^{\infty} \frac{te^{it}}{1+t^2} dt = \oint_{C} \frac{ze^{iz}}{1+z^2} dz = 2\pi i \operatorname{Res}(f,i) = \lim_{z \to i } 2\pi i \frac{ze^{iz}}{(z+i)} = \frac{\pi i}{e}$$ Therefore $$I = \frac{\pi}{2} - \Im(J) = \frac{\pi}{2} -\frac{1}{2}\Im \left(\frac{\pi i}{e}\right) = \frac{\pi}{2} - \frac{\pi}{2e} = \frac{\pi}{2}\left(1-\frac{1}{e}\right)$$ Hence, we can conclude $$\boxed{ \int_{0}^\infty \frac{\sin(\tan(x))}{x} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) }$$

Integral of the day XXIII

Nice integral solved with contour integration

Today we show the proof of this fun integral posted by @integralsbot $$\int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx = \frac{-1+2\sqrt{2}}{4}\pi -1$$ The proof will rely on contour integration round the unit complex circle.

Proof

$$\begin{align*} I = \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx =& \int_{0}^{1} \frac{1}{(1+\sqrt{1-w^2})(1+w^2)}dw \quad \left(w \mapsto \tan x \right)\\ =& \int_{0}^{1} \left( \frac{1-\sqrt{1-w^2}}{1-\sqrt{1-w^2}}\right)\frac{1}{(1+\sqrt{1-w^2})(1+w^2)}dw \\ =& \int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2(1+w^2)} dw\\ =& \underbrace{\int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2} dw}_{J} - \underbrace{\int_{0}^{1} \frac{1}{1+w^2} dw}_{K} + \underbrace{\int_{0}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw}_{L} \end{align*}$$ $$\begin{align*} J = \int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2} dw \stackrel{IBP}{=}& \frac{\sqrt{1-w^2}-1}{w}\Big|_{0}^{1} + 2\int_{0}^{1} \frac{1}{\sqrt{1-w^2}} dx \\ =& -1 + 2\arcsin(1)\\ =& -1 + \frac{\pi}{2} \end{align*}$$ $$\begin{align*} K = \int_{0}^{1} \frac{1}{1+w^2} dw = \arctan(1) = \frac{\pi}{4} \end{align*}$$ $$\begin{align*} L = \int_{0}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw =& \frac{1}{2} \int_{-1}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw \quad \textrm{(the integrand is even)}\\ =& \int_{0}^{\pi} \frac{\sin^2 s}{1+ \cos^2 s} ds \\ =& \int_{0}^{\pi} \frac{1-\cos^2 s}{1+ \cos^2 s} ds \\ =& \int_{0}^{\pi} \frac{2-(1+\cos^2 s)}{1+ \cos^2 s} ds \\ =& 2\int_{0}^{\pi} \frac{1}{1+ \cos^2 s} ds- \int_{0}^{\pi} ds\\ =& 4\int_{0}^{\pi} \frac{1}{3+ \cos(2s)} ds- \pi \quad \left( \cos^2 \vartheta = \frac{1+\cos(2\vartheta)}{2} \right)\\ =& 2\underbrace{ \int_{0}^{2\pi} \frac{1}{3+ \cos\theta}d\theta}_{M} - \pi \\ \end{align*}$$ To solve $M$ we can make the following change of variable $$\cos \theta = \frac{z+z^{-1}}{2}$$ $$d\theta = \frac{dz}{zi}$$ with this change the integral is transformed in a contour integral round the unit complex circle $$\begin{align*} M = \int_{0}^{2\pi} \frac{1}{3+ \cos\theta}d\theta = \frac{2}{i} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz \end{align*}$$ The only pole inside the unit circle is $z= 2\sqrt{2}-3$. Therefore by the residue theorem: $$\begin{align*} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz=& 2\pi i \operatorname{Res}\left(\frac{1}{z^2+6z+1},2\sqrt{2}-3\right) \\ =& \lim_{z \to 2\sqrt{2}-3} \frac{2\pi i}{z+3+2\sqrt{2}} \\ =& \frac{\pi i }{\sqrt{2}} \end{align*}$$ Hence $$M = \frac{2}{i} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz = \frac{2 \pi}{\sqrt{2}}$$ Therefore $$L = \frac{2}{\sqrt{2}}\pi - \pi$$ $$I = J-k+L = -1 + \frac{\pi}{2}- \frac{\pi}{4} + \frac{2}{\sqrt{2}}\pi - \pi = \frac{-1+2\sqrt{2}}{4}\pi -1$$ Finally, we can conclude $$\boxed{ \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx = \frac{-1+2\sqrt{2}}{4}\pi -1 }$$