# $\int_{0}^{\infty}{\sin^2bx\over x^2}\cdot{\mathrm dx\over e^{2ax}}=b\tan^{-1}\left({b\over a}\right)-{a\over 2}\ln{\left(a^2+b^2\over a^2\right)}?$

Consider $(1)$

$$\int_{0}^{\infty}{\sin^2bx\over x^2}\cdot{\mathrm dx\over e^{2ax}}=I\tag1$$ $(a,b)>0$

How does one show that $$I=\color{blue}{b\tan^{-1}\left({b\over a}\right)-{a\over 2}\ln{\left(a^2+b^2\over a^2\right)}}?$$

An attempt:

Using $\sin^2(x)={1-\cos(2x)\over 2}$

$(1)$ becomes

$${1\over 2}\int_{0}^{\infty}{e^{-2ax}\over x^2}-{e^{-2ax}\cos(2bx)\over x^2}{\mathrm dx}=I\tag2$$

$$I_1-I_2=I\tag3$$

$I_1$ is diverges, so else can we tackle $(1)?$

• Differentiate twice with respect to $b$, compute the result, integrate it twice. – Did Feb 21 '17 at 8:26
• Could you please change the title to something indicative of the question, or alternatively, of the topic of the question at hand? That way it would be more human-readable and can easily and concisely communicate what the question is about. – Aalok Feb 21 '17 at 9:50

Consider $\displaystyle F(a)=\int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx\\$

Double differentiating w.r.t 'a' we have,

$\displaystyle F''(a)=4\int\limits_0^\infty e^{-2ax}\sin^2 bx \; dx$

And a little integration by parts gives the last integral which simplifies as,

$\displaystyle F''(a)=\dfrac{1}{a}-\dfrac{a}{a^2+b^2}\\$

Now integrating , $\displaystyle F'(a)=\ln a-\dfrac{1}{2}\ln|a^2+b^2|+C_1$

Integrating for the last time we have,

$\displaystyle \int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx=b\arctan\left(\dfrac{a}{b}\right)-\dfrac{a}{2}\ln\left(\dfrac{a^2+b^2}{a^2}\right)+C_1a+C_2$

To eliminate the constants use $F'(1)=-\dfrac{1}{2}\ln|1+b^2|$ which will give $C_1=0$ and putting $b=0$ in the last equation will give $C_2=0$ and the result follows,

$\displaystyle \color{blue}{\int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx}=\color{red}{b\arctan\left(\dfrac{a}{b}\right)-\dfrac{a}{2}\ln\left(\dfrac{a^2+b^2}{a^2}\right)}$
