TITLE: Prove this bizarre integral:$\int_{0}^{\infty}{\sin(x^{\pi\over4})\over x[\cos(x^{\pi\over 4})+\cosh(x^{\pi \over 4})]}dx=1$
QUESTION [5 upvotes]: Integrate this bizarre integral,
$$I=\int_{0}^{\infty}{\sin(x^{\pi\over4})\over x[\cos(x^{\pi\over 4})+\cosh(x^{\pi \over 4})]}dx=1$$
I try:
Let generalise the integral and try and to determine the closed form,
$$\int_{0}^{\infty}{\sin(x^n)\over x[\cos(x^n)+\cosh(x^n)]}=F(n)$$
$$\cos{x}=\sum_{n=0}^{\infty}{(-1)^{2n}x^{2n}\over (2n)!}$$
$$\cosh{x}=\sum_{n=0}^{\infty}{x^{2n}\over (2n)!}$$
$$\cos{x}+\cosh{x}=2\left[{1+{x^4\over4!}+{x^8\over8!}+\cdots}\right]$$
Can't go any further, so I used  Wolfram integrator and try to figure out the closed form and I got $F(n)={\pi\over 4n}$.

REPLY [3 votes]: We may also use the identity $$\frac{\sin\left(x\right)}{\cosh\left(ax\right)+\cos\left(x\right)}=2\sum_{n\geq1}\left(-1\right)^{n-1}\sin\left(nx\right)e^{-anx},\, a>0,\, x\geq0.$$ We have $$ I=\int_{0}^{\infty}\frac{\sin\left(x^{b}\right)}{x\left(\cosh\left(x^{b}\right)+\cos\left(x^{b}\right)\right)}dx\stackrel{x^{b}=u}{=}\frac{1}{b}\int_{0}^{\infty}\frac{\sin\left(u\right)}{u\left(\cosh\left(u\right)+\cos\left(u\right)\right)}du.$$ Now fix $a>0$ and consider $$J\left(a\right)=\int_{0}^{\infty}\frac{u^{a-1}\sin\left(u\right)}{\cosh\left(u\right)+\cos\left(u\right)}du
 $$ $$=\frac{2}{b}\sum_{n\geq1}\left(-1\right)^{n-1}\int_{0}^{\infty}u^{a-1}\sin\left(nu\right)e^{-nu}du=\frac{2}{b}\textrm{Im}\left(\sum_{n\geq1}\left(-1\right)^{n-1}\int_{0}^{\infty}u^{a-1}e^{-nu\left(1-i\right)}du\right)
 $$ and now we can note that the last integral is the Gamma function. Hence $$J\left(a\right)=\frac{2}{b}\textrm{Im}\left(\frac{\Gamma\left(a\right)}{\left(1-i\right)^{a}}\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n^{a}}\right)=-\frac{2}{b}\Gamma\left(a\right)\left(1-2^{1-a}\right)\zeta\left(a\right)2^{a/2}\sin\left(\frac{\pi a}{4}\right)
 $$ hence $$I=\lim_{a\rightarrow0^{+}}J\left(a\right)=\color{red}{\frac{\pi}{4b}}$$ as wanted.