Quite complicated integral= sin(π/12)

[Tony1]

Given that,

How to show that,

$$\int_{0}^{\infty}{\cos(\pi x^2)\over {1\over 2}+\cosh\left({x\pi\over \sqrt{3}}\right)}\mathrm dx=\sin\left(\pi\over 12\right)$$

 

[jvicens]

Hello!

To show that the given integral is equal to $\sin\left(\frac{\pi}{12}\right)$, we can use the substitution $u=\frac{x}{\sqrt{3}}$ to rewrite the integral as:

$$\int_{0}^{\infty}{\cos(\pi x^2)\over {1\over 2}+\cosh\left({x\pi\over \sqrt{3}}\right)}\mathrm dx=\sqrt{3}\int_{0}^{\infty}{\cos\left(\pi\frac{u^2}{3}\right)\over {1\over 2}+\cosh(u\pi)}\mathrm du$$

Next, we can use the identity $\cosh(x)=\frac{e^x+e^{-x}}{2}$ to simplify the denominator:

$$\int_{0}^{\infty}{\cos\left(\pi\frac{u^2}{3}\right)\over {1\over 2}+\cosh(u\pi)}\mathrm du=\sqrt{3}\int_{0}^{\infty}{\cos\left(\pi\frac{u^2}{3}\right)\over \frac{1}{2}+\frac{e^{u\pi}+e^{-u\pi}}{2}}\mathrm du$$

Using the trigonometric identity $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, we can rewrite the numerator as:

$$\cos\left(\pi\frac{u^2}{3}\right)=\frac{e^{i\pi\frac{u^2}{3}}+e^{-i\pi\frac{u^2}{3}}}{2}$$

Substituting this into the integral, we get:

$$\sqrt{3}\int_{0}^{\infty}{\frac{e^{i\pi\frac{u^2}{3}}+e^{-i\pi\frac{u^2}{3}}}{2}\over \frac{1}{2}+\frac{e^{u\pi}+e^{-u\pi}}{2}}\mathrm du$$

Using the property of complex conjugates, we can rewrite the integral as:

$$\sqrt{3}\int_{0}^{\infty}{