When you ask my mentor : Am I any good at integrals ? You usually get an answer like this :

Give 5 proofs for $\int_0^{2 \pi} \ln( \frac{25}{16} - \sin(x)^2) dx = 0$.

I was able to show it with contour integration.

But there must be at least 4 other ways apparantly.

Can you show me them ?

In particular I wonder about parametric integration.

Example of parametric integration :

Let $y>0$ and

$$f(y) = \int_0^{oo} e^{-yx} \frac{sin(x)}{x} dx$$.

Then take the derivative with respect to $y$.

$$f ' (y) = - \int_0^{oo} e^{-yx} sin(x) dx = \frac{-1}{y^2 + 1}$$.

Integrating and taking into account $\lim_{y= oo} f(y) = 0$ we get

$$ f(y) = - \arctan(y) + \frac{\pi}{2} $$.


$$ \int_0^{oo} e^{-x} \frac{sin(x)}{x} dx = f(1) = \frac{\pi}{4} $$

Forgive me I do not have a formal definition for parametric integration but this example is useful I assume.

Closely related to parametric integration is the so-called derivative under the integral sign.

Ps : I forgot how I did the contour integral proof. Although I asked for the other 4 ways , you may give the contour method too. But that - alone - Will not be accepted as THE answer.

  • 9
    $\begingroup$ Some typo? What is inside your logarithm is negative. If one interpret it as a complex logarithm, one has a positive real part over the full interval, so the integral can't be zero. $\endgroup$ – mickep Nov 7 '15 at 12:31
  • 3
    $\begingroup$ This question is deeply flawed. See mickep's comment above and consider that $$\int_{0}^{2\pi}\log\left(\frac{625}{256}-\sin^4(x)\right)\,dx = 2\pi\cdot \log\frac{33+5\sqrt{41}}{32}.$$ $\endgroup$ – Jack D'Aurizio Nov 7 '15 at 12:53
  • 2
    $\begingroup$ Could you give us an outline of how you showed it with contour integration? If your proof is correct that would clear up some ambiguities, show the statement is true, and show that you have put significant work into the problem. If you don't do this, your question will probably be closed soon. $\endgroup$ – Rory Daulton Nov 7 '15 at 15:10
  • 3
    $\begingroup$ You still have the problem with negative logarithm. It should be $\log\left(\frac{25}{16}-\sin^2(x)\right)$ (at least that makes it correct) $\endgroup$ – Kibble Nov 7 '15 at 19:08
  • 1
    $\begingroup$ I voted for reopening, but think the question would be better if you showed the way you did it using contour integration. Also, maybe you should explain what you mean by "parametric integration"? Introducing a parameter? Write it as an curve integral (where one usually speaks about the parameter)? ... ? $\endgroup$ – mickep Nov 7 '15 at 19:09

Here is one way, using a parameter:

First, we note, by symmetry, $$ \int_0^{2\pi}\ln(25/16-\sin^2 x)\,dx=4\int_0^{\pi/2}\ln(25/16-\sin^2 x)\,dx. $$ Introduce, for $1\leq a$ $$ f(a)=4\int_0^{\pi/2}\ln(a^2-\sin^2x)\,dx. $$ Our task is to show that $f(5/4)=0$.

By symmetry and using the Euler log sin integral (see here for example) $$ \begin{split} f(1)&=4\int_0^{\pi/2}\ln(1-\sin^2x)\,dx\\ &=8\int_0^{\pi/2}\ln(\cos x)\,dx\\ &=8\int_0^{\pi/2}\ln(\sin x)\,dx\\ &=-4\pi\ln 2. \end{split} $$ Differentiating, we get, for $a>1$, $$ \begin{split} f'(a)&=8a\int_0^{\pi/2}\frac{1}{a^2-\sin^2x}\,dx\quad[u=\tan x]\\ &=\frac{8}{\sqrt{a^2-1}}\Bigl[\arctan\Bigl(\frac{\sqrt{a^2-1}}{a}u\Bigr)\Bigr]_0^{+\infty}\\ &=\frac{4\pi}{\sqrt{a^2-1}}. \end{split} $$ Now, $$ f(5/4)=f(1)+\int_1^{5/4}f'(a)\,da=-4\pi\ln 2+4\pi\Bigl[\ln(a+\sqrt{a^2-1})\Bigr]_1^{5/4}=0. $$

  • $\begingroup$ How did you do the last integral with the sine in the denominator ? $\endgroup$ – mick Nov 8 '15 at 20:21
  • $\begingroup$ If you add the link in the answer that might get a vote :) thanks. $\endgroup$ – mick Nov 8 '15 at 21:08
  • $\begingroup$ I understand , it just that Some people do not read comments ... That is their problem though. Anyways I assumed a very short way to do that integral - implied by getting the impression it was very trivial , too trivial too mention - but appar it is as hard as i expected. However i think i might be able to do it faster ... Need to consider that. $\endgroup$ – mick Nov 8 '15 at 21:19

One way: split the integral to the intervals $\left[0,\pi\right]$ and $\left[\pi,2\pi\right]$, and substitute $u=2\pi-x$ in the second integral.

  • $\begingroup$ How does this help? $\endgroup$ – Barry Cipra Nov 7 '15 at 20:46
  • $\begingroup$ Doesn't that substitution just give the same thing back, i.e. what you get is $2\int_0^\pi\ln(25/16-\sin^2(x))\,dx$. Since $\sin(2\pi-x)=-\sin x$, and the sine is squared, this does not help. $\endgroup$ – mickep Nov 7 '15 at 20:46

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.