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

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 $F(n)={\pi\over 4n}$.

Answer 1

After a change of variables $x^n=y$ (we assume $n>0$) the generalized integral reads

$$ F(n)=\frac{1}{n}\int_{0}^{\infty}\frac1y\frac{\sin(y)}{\cos(y)+\cosh(y)}=\frac{F(1)}{n} $$

so it is enough to evaluate the integral for $n=1$.

This may done via the residue theorem. Using parity we write

$$ F(1)=\frac12\Im\left[P\int_{\mathbb R}\frac{e^{iy}}{y(\cos(y)+\cosh(y))}\right] $$

the integrand nicely converges in the upper half of the complex plane and has poles at $z_0=0$ and $z_{k,\pm}=(\pm1+i)(\frac{\pi}{2}+2 \pi k)$ in this domain. It is now not too diffcult to show that $\text{Res}(z_{k+})+\text{Res}(z_{k-})=0$ for any $k\in \mathbb{N}$.

From the above considerations (the factor of $\pi i$ =$2 \pi i/2$ is due to the fact that the relevant singularity is on the contour of integration) we get

$$ F(1)=\frac{1}{2} \Im[\pi i \text{Res}(z_0)]=\frac{\pi}{4} $$

or

$$ F(n)=\frac{F(1)}{n}=\frac{\pi}{4 n} $$

putting $n=\frac{\pi}{4}$ we get $I=1$ as expected from OP's considerations

Answer 2

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.