Integrate $\int_0^{\infty} \frac{\sin^2 x}{\cosh x\>+\>\cos x}\frac{dx}x $

Question

It is known that (see for example) \begin{align} &\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4\\ &\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x =\frac\pi8 \end{align}

I am wondering if the similar integral below $$\int_0^{\infty} \frac{\sin^2 x}{\cosh x+\cos x}\frac{dx}x $$ can be evaluated to a simple close-form as well. I have tried the same approaches for the known integrals above and they do not lead to any results. Alternatively, I have also manipulated the integral and expressed it in the equivalent form $$\int_0^\infty e^{-x}\cos x\tanh x \>\frac{dx}x $$ which, though appearing simpler, is not any easier.

Answer 1

I'm not sure how you showed the two integrals are equivalent, but the following is an evaluation of $$\int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x}.$$

For $\Re(s) >0$, we have

$$ \begin{align} \int_{0}^{\infty} \tanh(t) e^{-st} \, \mathrm dt &= \int_{0}^{\infty} \frac{1-e^{-2t}}{1+e^{-2t}} \, e^{-st} \, \mathrm dt \\ &= \int_{0}^{\infty} (1-e^{-2t})e^{-st} \sum_{n=0}^{\infty} (-1)^{n}e^{-2tn} \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n}\int_{0}^{\infty}\left(e^{-(2n+s)t} -e^{-(2n+s+2)t} \right) \, \mathrm dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s+2} \\ &= \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}+1} \right) \\ &\overset{(1)}{=} \frac{1}{4} \left(\psi \left(\frac{s}{4} + \frac{1}{2}\right)- \psi \left( \frac{s}{4}\right) - \psi \left(\frac{s}{4}+1 \right) + \psi \left(\frac{s}{4}+\frac{1}{2} \right)\right) \\ &\overset{(2)}{=} \frac{1}{2} \left( \psi \left(\frac{s}{4}+ \frac{1}{2} \right) - \psi \left(\frac{s}{4} \right) - \frac{2}{s} \right). \end{align}$$

Therefore, $ \begin{align} \int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x} &= \Re\int_{0}^{\infty} e^{-(1+i)x} \frac{\tanh (x)}{x} \, \mathrm dx \\ &= \Re \int_{0}^{\infty}e^{-(1+i)x} \tanh(x) \int_{0}^{\infty} e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \Re \int_{0}^{\infty} \int_{0}^{\infty}\tanh(x) e^{-(1+i+t)x} \, \mathrm dx \, \mathrm dt \\ &= \Re \int_{0}^{\infty} \frac{1}{2}\left(\psi \left(\frac{1+i+t}{4}+ \frac{1}{2} \right) - \psi \left(\frac{1+i+t}{4} \right) - \frac{2}{1+i+t} \right) \, \mathrm dt \\ &= \Re \left(2 \ln\Gamma \left(\frac{3+i+t}{4} \right) -2 \ln \Gamma \left(\frac{1+i+t}{4} \right) -\ln (1+i+t)\right) \Bigg|_{0}^{\infty} \\ &\overset{(3)}= \Re \left(-2 \ln(2) - 2 \ln \Gamma \left(\frac{3+i}{4} \right)+2 \ln \Gamma \left(\frac{1+i}{4} \right) + \ln(1+i)\right)\\ &= - \frac{3 \ln (2)}{2} + 2\Re \left(\ln \Gamma \left(\frac{1+i}{4} \right)- \ln \Gamma \left(\frac{3+i}{4} \right) \right). \end{align}$

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$(1)$ For $\Re(z) > 0$, $\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+z} = \frac{1}{2} \left(\psi \left(\frac{z+1}{2} \right) - \psi \left(\frac{z}{2} \right) \right).$

$(2)$ Recurrence relation of the digamma function

$(3)$ For $a >0$, $\ln \Gamma(z_{1}+ax)- \ln \Gamma(z_{2}+ax) \sim (z_{1}-z_{2}) \ln(ax) + \mathcal{O} \left(\frac{1}{x}\right)$ as $x \to \infty$.

Answer 2

Another approach:

Let $$I= \int_{0}^{\infty} \frac{\sin^{2}(x)}{\cosh(x) + \cos (x)} \frac{\mathrm dx}{x}.$$

Using the identity $$\frac{\sin (x)}{\cosh(x) + \cos(x)} = 2 \sum_{n=1}^{\infty} (-1)^{n+1} \sin(nx) e^{-nx} = 2 \sum_{n=\color{red}{0}}^{\infty} (-1)^{n+1} \sin(nx) e^{-nx}, $$ we have

$ \begin{align} I &= 2\int_{0}^{\infty} \frac{\sin(x)}{x}\sum_{n=0}^{\infty} (-1)^{n+1}\sin(nx)e^{-nx} \, \mathrm dx \\ &= 2\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{\infty} \frac{\sin(x) \sin(nx)}{x} e^{-nx} \, \mathrm dx \\ &= 2\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{\infty} \sin(x) \sin(nx) \int_{0}^{\infty} e^{-(n+t)x} \, \mathrm dt \, \mathrm dx \\ &=2\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{\infty} \int_{0}^{\infty} \sin(x) \sin (nx)e^{-(n+t)x} \, \mathrm dx \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{\infty}\int_{0}^{\infty} \left(\cos\left((n-1)x \right)- \cos \left((n+1)x \right) \right)e^{-(n+t)x} \, \mathrm dx \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{\infty} \left(\frac{n+t}{(n-1)^{2}+(n+t)^{2}}- \frac{n+t}{(n+1)^{2}+(n+t)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2}\sum_{n=0}^{\infty} (-1)^{n+1} \ln \left(\frac{2n^{2}+2n+1}{2n^{2}-2n+1} \right) \\ &\overset{(1)}{=} \frac{1}{2} \sum_{n=0}^{\infty} \ln \left(\frac{2(2n)^{2}-2(2n)+1}{2(2n)^2+2(2n)+1} \ \frac{2(2n+1)^{2}+2(2n+1)+1}{2(2n+1)^{2}-2(2n+1)+1}\right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \ln \left(\frac{8n^{2}-4n+1}{8n^{2}+4n+1} \ \frac{8n^{2}+12n+5}{8n^{2}+4n+1} \right) \\ &= \frac{1}{2} \ln \left( \prod_{n=0}^{\infty} \frac{(8n^{2}-4n+1)(8n^{2}+12n+5)}{(8n^{2}+4n+1)^{2}} \right) \\ &= \frac{1}{2} \ln \left( \lim_{N \to \infty} \prod_{n=0}^{N} \frac{(8n^{2}-4n+1)(8n^{2}+12n+5)}{(8n^{2}+4n+1)^{2}} \right) \\ &= \frac{1}{2} \ln \left(\lim_{N \to \infty} \prod_{n=0}^{N} \frac{\left(n- \frac{1+i}{4} \right)\left(n- \frac{1-i}{4} \right) \left(n+ \frac{3-i}{4} \right)\left(n+ \frac{3+i}{4} \right)}{\left(n+ \frac{1-i}{4} \right)^{2}\left(n+ \frac{1+i}{4} \right)^{2} } \right) \\ & \overset{(2)}{=} \small \frac{1}{2} \ln \left( \lim_{N \to \infty} \frac{\Gamma \left(N+1- \frac{1+i}{4} \right) \Gamma \left(N+1- \frac{1-i}{4} \right) \Gamma \left(N+ 1+\frac{3-i}{4} \right) \Gamma\left(N+1+ \frac{3+i}{4} \right)\Gamma^{2} \left(\frac{1-i}{4}\right) \Gamma^{2} \left(\frac{1+i}{4} \right)}{ \Gamma \left(- \frac{1+i}{4} \right) \Gamma \left(-\frac{1-i}{4} \right) \Gamma \left(\frac{3-i}{4} \right) \Gamma \left(\frac{3+i}{4} \right)\Gamma^{2} \left(N+ 1+\frac{1-i}{4} \right) \Gamma^{2}\left(N+1+ \frac{1 +i}{4} \right)} \right) \\ & \overset{(3)}{=} \frac{1}{2} \ln \left( \lim_{N \to \infty} \frac{\Gamma^{4}(N) \, N^{(3-i)/4} N^{(3+i)/4} N^{(7-i)/4} N^{(7+i)/4} \, \Gamma^{2} \left(\frac{1-i}{4} \right)\Gamma^{2} \left(\frac{1+i}{4} \right)}{ 8 \, \Gamma^{2} \left(\frac{3-i}{4} \right) \Gamma^{2} \left(\frac{3+i}{4} \right)\, \Gamma^{4}(N) \, N^{(5-i)/2} N^{(5+i)/2} } \right) \\&=\frac{1}{2} \ln \left( \lim_{N \to \infty} \frac{N^{5} \, \Gamma^{2} \left(\frac{1-i}{4} \right)\Gamma^{2} \left(\frac{1+i}{4} \right)}{ 8 \, \Gamma^{2} \left(\frac{3-i}{4} \right) \Gamma^{2} \left(\frac{3+i}{4} \right) N^{5}} \right) \\ &= \frac{1}{2} \ln \left( \frac{\Gamma^{2} \left(\frac{1-i}{4} \right)\Gamma^{2} \left(\frac{1+i}{4} \right)}{ 8 \, \Gamma^{2} \left(\frac{3-i}{4} \right) \Gamma^{2} \left(\frac{3+i}{4} \right)} \right) \\ &= - \frac{3}{2} \ln(2) + \ln \left(\frac{\Gamma\left(\frac{1-i}{4} \right)\Gamma\left(\frac{1+i}{4} \right)}{ \Gamma \left(\frac{3-i}{4} \right) \Gamma \left(\frac{3+i}{4} \right)} \right). \end{align}$

You can use the property $\overline{\Gamma(z)} = \Gamma(\overline{z})$ to show that this is equivalent to the result from my other answer.

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$(1)$ Combine adjacent terms.

$(2)$ $\prod_{n=0}^{N} \left( n+z \right) = \frac{\Gamma(N+1+z)}{\Gamma(z)}$ if $z$ is not zero or a negative integer

$(3)$ $\Gamma(x+z) \sim \Gamma(x) x^{z}$ as $x \to + \infty$