prove that : $\int^{\infty}_0 \frac{\sin\left({\frac{2}{\pi}x^2}\right)}{\sinh^2(2x)} dx =\frac{1}{8}$

10 minbeginner

Problem

Prove the following definite integral identity: $\int_0^\infty\frac{\sin\!\left(\frac{2}{\pi}x^2\right)}{\sinh^2(2x)}\,dx=\frac{1}{8}.$

Solution 1

we have :

$\frac{\sin\left({\frac{2}{\pi}x^2}\right)}{\sinh^2(2x)} = \frac{1}{2\pi}\int^{\infty}_0 \frac{\cos\left({\frac{2}{\pi}tx}\right)}{\cosh(x)(\cosh(t)+\cosh(x))} dt$

Therfore:

\begin{aligned} I &= \int_{0}^{\infty} \frac{\sin\left(\frac{2}{\pi}x^2\right)}{\sinh^2(2x)}\,dx \\[4pt] &= \frac{1}{2\pi} \int_{0}^{\infty}\int_{0}^{\infty} \frac{\cos\left(\frac{2}{\pi}tx\right)} {\cosh(x)\left(\cosh(t)+\cosh(x)\right)} \,dt\,dx \\[4pt] &= \frac{1}{4\pi} \int_{0}^{\infty}\int_{0}^{\infty} \frac{\cos\left(\frac{2}{\pi}tx\right)} {\cosh(t)+\cosh(x)} \left( \frac{1}{\cosh(x)}+\frac{1}{\cosh(t)} \right) \,dt\,dx \\[4pt] &= \frac{1}{8} \int_{0}^{\infty} \frac{1}{\cosh^2(x)}\,dx \\[4pt] &= \frac{1}{8}. \end{aligned}

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