# Evaluating $\int_{0}^{\infty}(\ln \tan^2 bx)/(a^2+x^2)\ dx$

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.

• Are you sure that it converges? Commented Jul 9, 2012 at 16:49
• @Mercy Absolutely! Commented Jul 10, 2012 at 4:53
• How do you show it converges? Commented Jul 10, 2012 at 7:30

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 $$I(a,b)=\lim_{n \to \infty}\int_{L_n}f(z)dz=\frac{\pi}{a}\ln\tanh(ab).$$
• If $a=0$ and $b\neq0$ then it diverges by comparison with $\frac{1}{x^2}$ on $[0,\epsilon]$. Commented Jul 10, 2012 at 15:15
• Of course! that's why I do not consider the case $a=0$. Commented Jul 10, 2012 at 15:18