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How would I prove

$$\displaystyle\int_0^\infty \frac{\ln(\tan^2 (ax))}{1+x^2}dx = \pi\ln(\tanh(a))?$$

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Are we to assume a>0 –  Ethan Jan 27 '13 at 3:27
    
Doesn't look like the indefinite integral has a elementary closed form. –  Joe Jan 27 '13 at 3:27
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See here: math.stackexchange.com/questions/285960/… (essentially the same thing, asked a few days ago) –  L. F. Jan 27 '13 at 3:49
    
$\ln y^2=2\ln y$ –  Lucian Nov 17 '13 at 23:26
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2 Answers 2

up vote 22 down vote accepted

Here is another solution:

We remark that

$$ \log\tan^{2}\theta = -4 \sum_{n \ \mathrm{odd}}^{\infty}\frac{1}{n}\cos 2n\theta \tag{1} $$

and

$$ \sum_{n \ \mathrm{odd}}^{\infty} \frac{1}{n} x^{n} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right). \tag{2} $$

Both are easily proved by using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ and the Taylor series of the logarithm. Also we note that

$$ \int_{0}^{\infty} \frac{t \sin t}{a^2 + t^2} \, dt = \frac{\pi}{2} e^{-|a|}. \tag{3}$$

Then

\begin{align*} \int_{0}^{\infty} \frac{\log\tan^2(ax)}{1+x^2} \, dx &= \int_{0}^{\infty} \log\tan^2(ax) \left( \int_{0}^{\infty} \sin t \, e^{-xt} \, dt \right) \, dx \\ &= \int_{0}^{\infty} \sin t \int_{0}^{\infty} e^{-tx} \log\tan^2(ax) \, dxdt \\ &= -4 \int_{0}^{\infty} \sin t \int_{0}^{\infty} \sum_{n \ \mathrm{odd}}^{\infty}\frac{1}{n} e^{-tx} \cos (2nax) \, dxdt \\ &= -4 \int_{0}^{\infty} \sin t \sum_{n \ \mathrm{odd}}^{\infty}\frac{1}{n} \frac{t}{4a^{2}n^{2} + t^{2}} \, dt \\ &= -4 \sum_{n \ \mathrm{odd}}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{t \sin t}{4a^{2}n^{2} + t^{2}} \, dt \\ &= -2\pi \sum_{n \ \mathrm{odd}}^{\infty} \frac{1}{n} e^{-2an} = \pi \log \left( \frac{1-e^{-2a}}{1+e^{-2a}} \right) \\ &= \pi \log (\tanh a). \end{align*}

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Very nice. Where did you learn these techniques? I have seen you give similar impressive integral computations on many other questions. –  Potato Jan 27 '13 at 4:02
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@Potato, Actually I learned these techniques by myself. But I think that many other people are already aware of these. –  sos440 Jan 27 '13 at 4:19
    
How did you learn them? Was there a book or paper that was especially useful? I've seen similar tricks ("backwards" conversion of terms to integrals, power series) used by a few physicists, so I wonder if these techniques are taught in books on mathematical methods for physics. –  Potato Jan 27 '13 at 4:24
    
Thank You so much! –  Integrals and Series Jan 27 '13 at 8:48
    
@sos440: this proof is good for a collection with very nice proofs (+1) :-) –  Chris's sis Jan 27 '13 at 9:20
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An idea with complex contour. Let us choose the path

$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{it}\,\,,\,0\leq t\leq \pi\}\,\,,\,0<<R\in\Bbb R$$

Take the function

$$f(z):=\frac{\operatorname{Log}(\tan^2az)}{1+z^2}=2\frac{\operatorname{Log}(\tan az)}{1+z^2}$$

Inside the domain enclosed by $\,C_R\,$ above, the function has the pole $\,z=i\,$ (note that at the poles of $\,\tan az\,$ the logarithmic function equals $\,0+\arg(\text{pole})\,$ , and since we're going to choose the branch along the cut from zero to $\,-i\infty \,$ , i.e. the negative y-axis all these give us zero, so we're left only with the zero of the denominator in the positive half complex plane:

$$Res_{z=i}(f)=2\lim_{z\to i}(z-i)\frac{\operatorname{Log}(\tan az)}{z^2+1}=2\frac{\operatorname{Log}(\tan ai)}{2i}=-i\log(\tanh a)$$

We also have that

$$\left|\int_{\gamma_R}2\frac{\operatorname{Log}(\tan az)}{1+z^2}dz\right|\leq2\frac{|\log|\tan az||}{1-R^2}R\pi\xrightarrow[R\to\infty]{}0$$

as using the form (with $\,z=x+yi\,\,,\,x,y\in\Bbb R\,\,,\,y>0\,$)

$$\tan az=\frac{e^{2aiz}-1}{e^{2aiz}+1}\Longrightarrow |\log|\tan az||\leq \left|\log\frac{1+e^{2iy}}{1-e^{2iy}}\right|\xrightarrow [y\to\infty]{}\log 1=0$$

Thus, we finally get by Cauchy's Theorem

$$2\pi i(-i\log(\tanh a))=2\pi\log(\tanh a)=\oint_{C_R} f(z)\,dz=$$

$$=\int\limits_{-R}^R\frac{\log(\tan^2 ax)}{1+x^2}dx+\int_{\gamma_R}f(z)\,dz\xrightarrow[R\to\infty]{}\int\limits_{-\infty}^\infty\frac{\log(\tan^2 x)}{1+x^2}dx$$

Now just divide by two the integral of the even function above and we're done.

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"Infinitely many branch...*points*...? Along the real axis? No and no to both your questions. Read here in "branch cuts" and around: en.wikipedia.org/wiki/Complex_logarithm#Branch_cuts –  DonAntonio Mar 13 '13 at 23:16
    
@RandomVariable, I think you don't quite know exactly what a branch cut is (not point). Google it, read it books, in the link I sent you...At the origin, $\log z\,$ isn't defined... but...that's not the matter: if you go "around" $\,z=0\,$ , the argument of complex numbers increases (or decreases, depending on the spinning direction) by an integer multiple of $\,2\pi\,$, and this makes the value of $\,\log z:= \log|z|+i\arg z\,$ multivalued, and from here that branch cutes are chosen to "prevent" that spinning and make $\,\log\,$ single valued...it's too messy to explain here. –  DonAntonio Mar 14 '13 at 2:10
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