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## A quick integral July 24, 2009

Posted by Akhil Mathew in Problem-solving.
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The integral ${I=\int_0^\pi \log \sin x dx}$ is normally computed (e.g. in Ahlfors’ book) to be ${-\pi \log 2}$ using complex integration over a suitable almost-rectangular contour. There is also a simple and direct way to get the value of this integral by a substitution and elementary calculus.

First, by the substitution ${x=2t}$ and the identity ${\sin(2x)=2\sin x \cos x}$,

$\displaystyle I = 2 \int_0^{\pi/2} \log \sin t dt + 2 \int_0^{\pi/2} \log \cos t dt + \pi \log 2;$

then using the symmetry of ${\sin}$ and ${\cos}$ gives:

$\displaystyle I = I + I + \pi \log 2,$

whence the result. There are slight technicalities regarding the improperness of these integrals, but they can be directly justified (or one may use the Lebesgue integral).

[Edit (7/25)- Todd Trimble posted solutions to similar integrals, which use the result of this post as a lemma, here.  AM]

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## Comments»

1. Qiaochu Yuan - July 24, 2009

There seem to be lots of neat ways to exploit symmetry to compute certain definite integrals; my favorite trick overall has to be the introduction of a parameter and differentiation under the integral sign. Apparently the theory of definite integrals, such as it exists, is not fully understood, although Doron Zeilberger has tried to place at least an important part of it on a firm footing.

2. Akhil Mathew - July 25, 2009

This is definitely a useful technique as well; I recall seeing it used to prove nontrivial results on definite integrals some time back with ease, although my memory’s not good enough to recall them at will. Perhaps this was something of the form
$\int_0^\infty \log(y/x) f(x) dx$? I will have to check, and perhaps post sometime later about them.

In fact, it’d be of interest later to discuss differentiation and integration sometime later in the more general setting of Banach spaces, and work these results out there with the generality.

3. Richard - July 25, 2009

There’s also a fun way of getting this integral by finding $\latex \prod \sin(k \pi/n)$ and taking the limit $n \rightarrow \infty.$

4. Todd Trimble - July 25, 2009

Incidentally, this problem also came up on our blog; some solutions are given here, where some discussion of differentiation under the integral sign was also given.

5. Akhil Mathew - July 25, 2009

I ought to have checked more carefully that I wasn’t repeating another math blog, at least for puzzle-like problems such as these, even though I was sure this was a well-known fact. In any case, I’ll add a link to your post. Thanks for pointing that out.

6. Todd Trimble - July 25, 2009

No problem (and you’re of course right that it’s a well-known fact) — I wasn’t fussed about it. I’m merely following up on Qiaochu’s comment, and indicating that it can be evaluated by a differentiation under the integral technique. Thanks for the link!

7. 两个快速积分 « Liu Xiaochuan’s Weblog - July 26, 2009

[…] 其一是$latex {I=int_0^pi log sin x dx}”$，首先令,注意到等式,于是 […]