35
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ Irrelevant to the question but: this came as a problem in a PhD entry test for what subject? $\endgroup$ – user258700 Jul 18 '16 at 18:17
  • 3
    $\begingroup$ @AhmedHussein My question too, who made you do this??? $\endgroup$ – qbert Jul 18 '16 at 18:18
  • 24
    $\begingroup$ "We want our PhDs to know all sorts of integral tricks! That is, after all, what mathematics is truly about." $\endgroup$ – parsiad Jul 18 '16 at 18:27
  • 7
    $\begingroup$ Definitely not a pure math PhD program. $\endgroup$ – Mathemagician1234 Jul 18 '16 at 18:35
  • 4
    $\begingroup$ 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. $\endgroup$ – abnry Jul 18 '16 at 18:38
25
$\begingroup$

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:

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

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ Thanks for the answer and introducing gd function. (+1) $\endgroup$ – Venus Jul 18 '16 at 19:17
  • $\begingroup$ I corrected a small typo. The final expression using the Gudermannian function should have the factor $\frac12$ to evaluate as $I \approx 7.11363$. $\endgroup$ – Tito Piezas III May 19 at 3:56
  • $\begingroup$ @TitoPiezasIII: appreciated, thank you. $\endgroup$ – Jack D'Aurizio May 19 at 21:13
17
$\begingroup$

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 $(2)$.

|cite|improve this answer
$\endgroup$
  • 2
    $\begingroup$ very nice....(+1) $\endgroup$ – tired Jul 18 '16 at 18:48
  • $\begingroup$ It seems very promising, let me take a look. (+1) $\endgroup$ – Venus Jul 18 '16 at 18:48
  • 6
    $\begingroup$ call me stupid, but how exactly do we get (3) ? $\endgroup$ – tired Jul 18 '16 at 18:56
  • 7
    $\begingroup$ 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 $\sin (\tau x)$. $\endgroup$ – Olivier Oloa Jul 18 '16 at 19:00
  • 7
    $\begingroup$ So the final result is? $\endgroup$ – marty cohen Jul 18 '16 at 19:11
11
$\begingroup$

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...

|cite|improve this answer
$\endgroup$
  • 13
    $\begingroup$ Still busy playing Pokemon Go ? $\endgroup$ – A---B Nov 3 '17 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.