The entry-level PhD integral: $\int_0^\infty\frac{\sin 3x\sin 4x\sin5x\cos6x}{x\sin^2 x\cosh x}\ dx$

Question

I hope you find this integral interesting.

Evaluate $$\int_0^\infty\frac{\sin\left(\,3x\,\right)\sin\left(\,4x\,\right) \sin\left(\,5x\,\right)\cos\left(\,6x\,\right)}{x\,\sin^{2}\left(\,x\,\right)\cosh\left(\,x\,\right)}\,\,\mathrm{d}x\tag1$$

This problem is taken from the PhD graduate entry tests in my college. I've tried to use product-to-sum trigonometric identities $$2\sin 4x\sin 3x=\cos x-\cos 5x$$ and $$2\cos 6x\sin 5x=\sin 11x-\sin x$$ I got a bunch of the following form $$\int_0^\infty\frac{\sin \alpha x\cos \beta x}{x\sin^2 x\cosh x}\ dx\quad\Longrightarrow\quad\int_0^\infty\frac{\sin \gamma x}{x\sin^2 x\cosh x}\ dx\tag2$$ I tried $$I'(\gamma)=\int_0^\infty\frac{\cos \gamma x}{\sin^2 x\cosh x}\ dx\tag3$$ but the latter form is not easy to evaluate either. Can anyone here help me to evaluate $(1)$? Thanks in advance.

Answer 1

Hint. One may start with the standard evaluation $$ \int_0^\infty \frac{\cos (ax)}{\cosh x}\:dx=\frac{\pi}2\:\frac1{\cosh \left(\large \frac{\pi a}2\right)},\quad a\ge 0,\tag1 $$ then, writing $\displaystyle \frac1{2\cosh \left(\large \frac{\pi a}2\right)}=\frac{e^{\large \frac{\pi a}2}}{e^{a\pi}+1}$, integrating it with respect to $a$ from $0$ to $b$ gives $$ \int_0^\infty \frac{\sin (b x)}{x\cosh x}\:dx=2\arctan\left(\tanh\left(\frac{b \pi }{4}\right)\right). \tag2 $$ Now one may just observe that $$ \frac{\sin 3x\sin 4x\sin5x\cos6x}{x\sin^2 x\cosh x}=\sum_{\large b_i}\alpha_i\frac{\sin (b_i x)}{x\cosh x} \tag3 $$ and conclude using $(2)$.

Answer 2

By De Moivre's formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ we have the following Fourier sine series: $$\frac{\sin(3x)\sin(4x)\sin(5x)\cos(6x)}{\sin^2(x)}\\= -\frac{1}{2} \sin(2x)-\frac{1}{2}\sin(4x)+\sin(8x)+\frac{3}{2}\sin(10x)+\frac{3}{2}\sin(12x)+\sin(14x)+\frac{1}{2}\sin(16 x)$$ and: $$I(n)=\int_{0}^{+\infty}\frac{\sin(2nx)}{x\cosh(x)}\,dx = 2\arctan\left(\tanh\frac{\pi n}{2}\right) $$ follows by differentiation under the integral sign. The original integral can so be expressed in terms of the Gudermannian function:

$$ I = \frac{1}{2} \big(-\text{gd}(\pi)- \text{gd}(2\pi) + 2 \text{gd}(4\pi) + 3 \text{gd}(5\pi) + 3 \text{gd}(6\pi) + 2 \text{gd}(7\pi) + \text{gd}(8\pi)\big) \approx 7.11363 $$

Answer 3

In fact, we have $$ \begin{align} I(M,N)&=\int_0^\infty\frac{\sin Nx\sin(N+1)x\sin Mx\cos(M+1)x}{x\sin^2 x\cosh x}\ dx\\[10pt] &=\sum_{m=1}^M\sum_{n=1}^N\left[\arctan\left( e^{(m+n)\pi} \right)-\arctan\left( e^{(m-n)\pi} \right)\right]\\[10pt] &=\frac{1}{2}\sum_{m=1}^M\sum_{n=1}^N\bigg[\operatorname{gd}\!\big((m+n)\pi\big)-\operatorname{gd}\!\big((m-n)\pi\big)\bigg] \end{align} $$ and the desired integral is $I(5,3)$.

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_{Sorry for the Cleo-style answer but right now I'm busy playing Pokemon Go, so I'll post the complete solution when I'm free. See ya...}