Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$ \mbox{How would I prove}\quad \int_{0}^{\infty} {\ln\left(\,\tan^{2}\left(\, ax\,\right)\,\right) \over 1 + x^{2}}\,{\rm d}x =\pi \ln\left(\,\tanh\left(\,\left\vert\, a\,\right\vert\,\right)\,\right)\,{\Large ?}. \qquad a \in {\mathbb R}\verb*\* \left\{\,0\,\right\} $$

share|cite|improve this question
Doesn't look like the indefinite integral has a elementary closed form. – Joe Jan 27 '13 at 3:27
See here:… (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
@Ethan It should be $\displaystyle\large \pi\ln\left(\,\tanh\left(\,\left\vert\, a\,\right\vert\,\right)\,\right)\,,\quad a \not= 0$. – Felix Marin Nov 17 '14 at 5:02

4 Answers 4

up vote 32 down vote accepted

Here is another solution:

We remark that

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


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

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

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


\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*}

share|cite|improve this answer
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
@Potato, Actually I learned these techniques by myself. But I think that many other people are already aware of these. – Sangchul Lee 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 the artist Jan 27 '13 at 9:20

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.

share|cite|improve this answer
"Infinitely many branch...*points*...? Along the real axis? No and no to both your questions. Read here in "branch cuts" and around: – 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's too messy to explain here. – DonAntonio Mar 14 '13 at 2:10

Denote the evaluated integral as $I$, then $I$ may be rewritten as $$I=\int_0^\infty \frac{\ln \sin^2 ax}{1+x^2}\,dx-\int_0^\infty \frac{\ln \cos^2 ax}{1+x^2}\,dx$$ Using Fourier series representations of $\ln \sin^2 \theta$ and $\ln \cos^2 \theta$, $$\ln \sin^2 \theta=-2\ln2-2\sum_{k=1}^\infty \frac{\cos2k\theta}{k}$$ and $$\ln \cos^2 \theta=-2\ln2+2\sum_{k=1}^\infty (-1)^{k+1}\frac{\cos2k\theta}{k}$$ also note that $$\int_0^\infty\frac{\cos bx}{1+x^2}\,dx=\frac{\pi e^{-b}}{2}$$ then $$\begin{align}I&=-2\sum_{k=1}^\infty \frac{1}{k}\int_0^\infty\frac{\cos2kax}{1+x^2}\,dx-2\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\int_0^\infty\frac{\cos2kax}{1+x^2}\,dx\\&=-\pi\sum_{k=1}^\infty \frac{e^{-2ak}}{k}-\pi\sum_{k=1}^\infty (-1)^{k+1}\frac{e^{-2ak}}{k}\\&=\pi\ln\left(1-e^{-2a}\right)-\pi\ln\left(1+e^{-2a}\right)\\&=\pi\ln\left(\frac{1-e^{-2a}}{1+e^{-2a}}\right)\\&=\pi\ln\left(\tanh a\right)\end{align}$$

share|cite|improve this answer

By equating the real parts on both sides of the identity $$\sum_{k=1}^{\infty} \frac{(be^{i \theta })^{k}}{k} = -\log(1-be^{i \theta}) \ , \ |b| < 1,$$ one finds that $$ \sum_{k=1}^{\infty} \frac{b^{k} \cos k\theta }{k} = - \frac{1}{2} \log \left(1-2b \cos \theta +b^{2} \right).$$

Then for $a >0$, $$ \begin{align}\int_{0}^{\infty} \frac{\log(1-2b \cos 2ax +b^{2})}{1+x^{2}} \ dx &= -2\int_{0}^{\infty}\frac{1}{1+x^{2}}\sum_{k=1}^{\infty} \frac{b^{k} \cos (2akx)}{k} \\ &=-2 \sum_{k=1}^{\infty} \frac{b^{k}}{k}\int_{0}^{\infty} \frac{\cos (2ak x)}{1+x^{2}} \ dx \\ &=-\pi \sum_{k=1}^{\infty} \frac{(be^{-2a})^{k}}{k} \\ &= \pi \log(1-be^{-2a}). \end{align}$$

Letting $b \to 1^{-}$, $$ \int_{0}^{\infty} \frac{\log (4 \sin^{2} ax)}{1+x^{2}} \ dx = \pi \log(1-e^{-2a}). \tag{1}$$

While letting $b \to -1^{+}$, $$\int_{0}^{\infty} \frac{\log(4 \cos^{2} ax)}{1+x^{2}} \ dx = \pi\log(1+e^{-2a}). \tag{2}$$

Then subtracting $(2)$ from $(1)$, $$ \int_{0}^{\infty} \frac{\log (\tan^{2} ax)}{1+x^{2}} \ dx = \pi \log \left(\frac{1-e^{-2a}}{1+e^{-2a}} \right) =\pi \log( \tanh a) .$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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