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Some time ago I came across one of the integrals, which still goes over my mind:

$$\int_{0}^{\infty}\frac{\ln \tan^2(bx)}{a^2+x^2}dx$$ a and b are parameters.

I would be interested in possible solutions with complex analysis and without it as well.

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Are you sure that it converges? –  Mercy Jul 9 '12 at 16:49
    
@Mercy Absolutely! –  Martin Gales Jul 10 '12 at 4:53
    
How do you show it converges? –  Mercy Jul 10 '12 at 7:30
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1 Answer 1

up vote 7 down vote accepted

If one can prove that the given integral converges, it's not hard to compute its value. Let's assume from now that the integral does converge. Since $\tan^2(-bx)=\tan^2(bx)$ and $(-a)^2=a^2$, there is no loss of generality in assuming that $a,b>0$. Then $$ I(a,b)=\int_0^\infty\frac{\ln\tan^2(bx)}{a^2+x^2}dx=2b\int_0^\infty\frac{\ln|\tan x|}{a^2b^2+x^2}dx=b\int_\mathbb{R}\frac{\ln|\tan x|}{a^2b^2+x^2}dx. $$ Consider the function $$ f: \mathbb{C} \to \mathbb{C},\ f(z)=b\frac{\ln|\tan z|}{a^2b^2+z^2}. $$

Given $n \in \mathbb{N}$, with $0<1/n<ab<n$, we denote by $\Delta_n$ the bounded region of $\mathbb{C}$ whose boundary consists of the segment $$ L_n=\{ x-\frac{i}{n}:\ |x|\le n\pi\} $$ and the upper half circle $$ \Gamma_n=\{\gamma_n(t)=-\frac{i}{n}+(n+\frac{1}{8})\pi e^{it}: \ 0 \le t \le \pi\}. $$

The set of poles of $f$ that lie inside $\Delta_n$, is $P=\{iab, k\pi/2:\ |k|\le 2n\}$.

For every $k$ with $|k|\le 2n$, $z_k=k\pi/2$ is a pole of order 2 with $$ \text{Res}(f,z_k)=\lim_{z \to 0}\frac{d}{dz}(z^2f(z+z_k))=0, $$ and since $$ \text{Res}(f,iab)=\frac{1}{2ia}\ln\tanh(ab), $$ we have $$ \int_{\Delta_n}f(z)dz=i2\pi\text{Res}(f,iab)=\frac{\pi}{a}\ln\tanh(ab). $$ Hence $$ \int_{L_n}f(z)dz=\frac{\pi}{a}\ln\tanh(ab)-J_n $$ with $$ J_n:=\frac{\pi}{a}\ln\tanh(ab)-in\pi\int_0^\pi e^{it}f((n+\frac{1}{8})\pi e^{it}-\frac{i}{n})dt. $$ Notice that \begin{eqnarray} |J_n|&\le&(n+\frac{1}{8})\pi\int_0^\pi|f((n+\frac{1}{8})\pi e^{it}-\frac{i}{n})|dt\cr &\le& \frac{(n+\frac{1}{8})\pi}{((n+\frac{1}{8})\pi-\frac{1}{n})^2-a^2b^2}\int_0^\pi|\ln|\tan((n+\frac{1}{8})\pi e^{it}-\frac{i}{n})||dt\cr &=&\frac{(n+\frac{1}{8})\pi}{((n+\frac{1}{8})\pi-1/n)^2-a^2b^2}\int_0^\pi\left|\ln\left|\frac{\exp(i(2n+\frac{1}{4})\pi e^{it}+\frac{2}{n})-1}{\exp(i(2n+\frac{1}{4})\pi e^{it}+\frac{2}{n})+1}\right|\right|dt\cr &\le&\frac{(n+\frac{1}{8})\pi}{((n+\frac{1}{8})\pi-\frac{1}{n})^2-a^2b^2}A_n, \end{eqnarray} with $$ A_n=\int_0^\pi |\ln|e^{i(2n+\frac{1}{4})\pi\cos t}e^{\frac{2}{n}-(2n+\frac{1}{4})\pi\sin t}-1|+|\ln|e^{i(2n+\frac{1}{4})\pi\cos t}e^{-(2n+\frac{1}{4})\pi\sin t+\frac{2}{n}}+1||dt. $$ $A_n$ is clearly bounded, so we conclude that $J_n \to 0$ as $n \to \infty$, and $$ I(a,b)=\lim_{n \to \infty}\int_{L_n}f(z)dz=\frac{\pi}{a}\ln\tanh(ab). $$

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Very nice! So you agree that the integral converges? –  Martin Gales Jul 10 '12 at 13:52
    
If $a=0$ and $b\neq0$ then it diverges by comparison with $\frac{1}{x^2}$ on $[0,\epsilon]$. –  nullUser Jul 10 '12 at 15:15
    
Of course! that's why I do not consider the case $a=0$. –  Mercy Jul 10 '12 at 15:18
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