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

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.

• Irrelevant to the question but: this came as a problem in a PhD entry test for what subject?
– user258700
Jul 18, 2016 at 18:17
• @AhmedHussein My question too, who made you do this??? Jul 18, 2016 at 18:18
• "We want our PhDs to know all sorts of integral tricks! That is, after all, what mathematics is truly about." Jul 18, 2016 at 18:27
• Definitely not a pure math PhD program. Jul 18, 2016 at 18:35
• I mean, integral tricks are actually useful to know, especially in analysis and such. But this integral just looks tedious, regardless of the tricks you use. That's my issue. Jul 18, 2016 at 18:38

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:

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

• Thanks for the answer and introducing gd function. (+1) Jul 18, 2016 at 19:17
• I corrected a small typo. The final expression using the Gudermannian function should have the factor $\frac12$ to evaluate as $I \approx 7.11363$. May 19, 2019 at 3:56
• @TitoPiezasIII: appreciated, thank you. May 19, 2019 at 21:13
• $+1$ thanks for introducing "Gudermannian function"
– user960916
Oct 9, 2021 at 18:15

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 $(2)$.

• very nice....(+1) Jul 18, 2016 at 18:48
• It seems very promising, let me take a look. (+1) Jul 18, 2016 at 18:48
• call me stupid, but how exactly do we get (3) ? Jul 18, 2016 at 18:56
• From $\sin(3x)\sin(4x)$ one gets rid of the factor $\sin^2 x$ in the denominator. Then we have a sum of $\sin (\alpha x) \cos (\beta x)$ which ends up as a linear sum of $\sin (\tau x)$. Jul 18, 2016 at 19:00
• So the final result is? Jul 18, 2016 at 19:11

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 $I(5,3)$.

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...

• Still busy playing Pokemon Go ?
– user312097
Nov 3, 2017 at 18:20
• Taking procrastination to a whole other level. Nov 7, 2019 at 14:06