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Integrate $\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx$

$\newcommand{\intd}[1]{\,\mathrm{d}#1}$ My Attempt

Put $t=\tan x\implies\intd t=\sec^2x\intd x$ \begin{align*} \int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx&=\int_0^{\pi/2}\frac{\tan x\cdot\sec^2x}{(1+m^2\tan^2 x)\sec^2x}\intd x\\ &=\int_0^{\infty}\frac{t}{(1+m^2t^2)(1+t^2)}\intd t \end{align*} Put $t^2=y\implies 2t\intd t=\intd y$ \begin{align*} \frac{1}{2}\int_0^\infty\frac{dy}{(1+m^2y)(1+y)}&=\frac{1}{2}\int_0^\infty\bigg[\frac{1}{1-m^2}\cdot\frac{1}{1+y}+\frac{m^2}{m^2-1}\cdot\frac{1}{1+m^2y}\bigg]dy\\ &=\frac{1}{2(1-m^2)}\int_0^\infty\frac{dy}{1+y}-\frac{m^2}{2(1-m^2)}\int_0^\infty\frac{dy}{1+m^2y}\\ &=\bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty\\ &=\bigg[\frac{1}{2(1-m^2)}\log|\frac{1+y}{1+m^2y}|\bigg]_0^\infty=\color{red}{\frac{1}{2(1-m^2)}\bigg[\frac{\infty}{\infty}-0\bigg]}\end{align*} I think I am getting stuck here as the substitutions does not seem to give the solution ?

Doubt $$ \lim_{y\to 0}\log|\frac{1+y}{1+m^2y}|=\log 1=0\\ \lim_{y\to \infty}\log|\frac{1+y}{1+m^2y}|=\lim_{y\to \infty}\log|\frac{\frac{1}{y}+1}{\frac{1}{y}+m^2}|=\log\frac{1}{m^2} $$ So what really we are doing with definite integrals ?. Does this mean that we are actually computing the upper and lower limits and taking the difference to find the definite integral ?

Note: Expanding in terms of $\sin x$ and $\cos x$ gives the solution, no doubt about limit going to infinity, not looking for that.

Sooraj S
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  • https://math.stackexchange.com/questions/1005976/finding-int-0-pi-2-frac-tan-x1m2-tan2x-mathrmdx/1006068#1006068 – tired Jun 05 '18 at 07:12
  • @tired how do u find example qstn. I searched, nothing showed up. btw i have edited OP, hope it makes my doubt clear. – Sooraj S Jun 05 '18 at 07:28
  • i rembered that i answered something similar many years ago :) – tired Jun 05 '18 at 21:16

2 Answers2

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You have one error $$\frac{m^2}{1-m^2}\int\frac{1}{1+m^2y}\,\mathrm{d}y=\frac{1}{1-m^2}\log|1+m^2y|+C$$ So your final answer is \begin{align*} \bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty&=\frac{1}{2(1-m^2)}\left[\log|1+y|-\log|1+m^2y|\right]_0^\infty\\ &=\frac{1}{2(1-m^2)}\log\left|\frac{1+y}{1+m^2y}\right|\bigg\rvert_0^\infty\\ &=\frac{1}{2(1-m^2)}\log\left(\frac1{m^2}\right)\\ &=\boxed{\frac{\log m}{m^2-1}} \end{align*}

Teh Rod
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  • thnx. it was by mistake.. my actual doubt was regarding the $\frac{\infty}{\infty}-0$ form in the solution – Sooraj S Jun 05 '18 at 07:02
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I think that you could have done it faster avoiding the second change of variable since, as you wrote, $$ I=\int\frac{\tan (x)}{1+m^2\tan^2 (x)}\,dx=\int\frac{t}{(1+m^2t^2)(1+t^2)}\,dt $$ Now, partial fraction decomposition gives $$\frac{t}{(1+m^2t^2)(1+t^2)}=\frac{m^2 t}{\left(m^2-1\right) \left(1+m^2 t^2\right)}-\frac{t}{\left(m^2-1\right) \left(1+t^2\right)}$$ making $$I=\int\frac{t}{(1+m^2t^2)(1+t^2)}\,dt=\frac{\log \left(1+m^2 t^2\right)-\log \left(1+t^2\right)}{2(m^2-1)}$$ $$I=\frac 1{2(m^2-1)}\log\left(\frac{1+m^2t^2}{1+t^2}\right)$$

Claude Leibovici
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