Take the 2-minute tour ×
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.

How to prove that

$$\operatorname{P.V.}\int^\infty_{-\infty}\frac{\ln(t^2+1)}{t^2-1}dt =\frac{\pi^2}{2}\: ?$$

share|improve this question
5  
Please say where the problem is from, and if it is homework please tag it as such. Welcome to math.SE! –  nbubis Nov 1 '12 at 11:57
add comment

2 Answers

up vote 6 down vote accepted

If you integrate along this closed curve in $\mathbb{C}$

enter image description here

then you get $0$ by Cauchy theorem. The integral you want is along the horizontal parts of the contour, as the small circles vanish and the big one goes to $\infty$. The contributions of the small circles cancel each other (the two residues have opposite signs), the integral along to big semicircle goes to $0$, so your integral is equal to the integral along the vertical part of the contour. Now notice that $\log(t^2+1)$ have different values on the two sides of the vertical path (that's why the integral is not $0$), but they differ just by $2\pi i$. What remains is (setting $t=is$) $$\int_1^\infty 2\pi/(1+s^2)\, ds=\pi^2/2$$

share|improve this answer
add comment

Since the integrand is even,

$$ \begin{align*} PV \int_{-\infty}^{\infty} \frac{\log(t^2+1)}{t^2 - 1}\,dt &= 2PV\int_{0}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt \\ &= 2PV \lim_{\epsilon \to 0+} \left( \int_{0}^{1-\epsilon} \frac{\log(t^2+1)}{t^2-1} \, dt + \int_{1+\epsilon}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt \right) \end{align*} $$

By the substitution $t \mapsto t^{-1}$, we have

$$\int_{1+\epsilon}^{\infty} \frac{\log(t^2+1)}{t^2-1} \, dt = \int_{0}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1) - 2\log t}{1-t^2} \, dt.$$

Thus we obtain

$$PV \int_{-\infty}^{\infty} \frac{\log(t^2+1)}{t^2 - 1}\,dt = 2PV \lim_{\epsilon \to 0+} \left( \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1)}{1-t^2} \, dt - 2\int_{0}^{\frac{1}{1+\epsilon}} \frac{\log t}{1-t^2} \, dt \right).$$

One the one hand, the function $\frac{1}{t+1}\log(t^2+1)$ is continuous on $[0, 1]$, hence is bounded on this interval by some constant $C > 0$. Also, if $\epsilon > $ is sufficiently small, we have $1 - \epsilon < \frac{1}{1+\epsilon}$. Thus we obtain an estimate

$$ \left| \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{\log(t^2+1)}{1-t^2} \, dt \right| \leq C \int_{1-\epsilon}^{\frac{1}{1+\epsilon}} \frac{dt}{1-t} = C \log \left(1+\epsilon\right).$$

This proves that the left-hand side converges to 0 as $\epsilon \to 0+$. On the other hand, by Tonell's theorem

$$ \begin{align*} \lim_{\epsilon \to 0+} \int_{0}^{\frac{1}{1+\epsilon}} \frac{\log t}{1-t^2} \, dt &= \int_{0}^{1} \frac{\log t}{1-t^2} \, dt = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \, dt \\ &= -\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = -\frac{\pi^2}{8}, \end{align*}$$

where at the last equality we used the famous Euler's series. Therefore the desired result follows.

share|improve this answer
add comment

Your Answer

 
discard

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.