Integrate $I=\int_0^\infty \frac{e^{-ax}\sin bx}{x}\mathrm dx$ If $a > 0$ then show
$$\int_0^\infty \frac{e^{-ax}\sin bx}{x} \mathrm dx=\tan^{-1}\frac{b}{a},$$
hence also show that
$$(i) \quad \int_0^\infty \frac{\sin bx}{x}\mathrm dx=\frac{\pi}{2} \text{ when } b>0,$$
$$(ii) \quad\int_0^\infty \frac{\sin bx}{x}\mathrm dx=-\frac{\pi}{2} \text{ when } b<0.$$
$$I=\int_0^\infty \frac{e^{-ax}\sin bx}{x}\mathrm dx .$$
$$\frac{dI}{db}=\int_0^\infty \frac{e^{-ax}}{x} \frac{\partial}{\partial b} (\sin bx)\mathrm dx 
 =\int_0^\infty \frac{e^{-ax}}{x}x\cos bx\mathrm dx 
 =\frac{a}{a^2+b^2}.$$
But, how is $$e \int_0^\infty \frac{e^{-ax}}{x}x\cos bx\mathrm dx = \frac{a}{a^2 +b^2}.$$ I was solving the problem by differentiating inside the integral sign. There's an $x$ in denominator. So, It's not my answer.
Note: I am doing differentiation under integral sign.
Question: Show that $\int_0^\infty e^{-ax}\cos bx\mathrm dx=\frac{a}{a^2 +b^2}$ (recent linked question didn't prove it. They were just telling to prove what I will get if I integrate by parts. And, another one prove my question is equal to 1/2. Thats why I am rejecting them. There's no possible answer which matches with mine.)
 A: Well, we are trying to find:
$$\mathcal{I}\left(\sigma\space;\beta\right):=\int_0^\infty\frac{\exp\left(-\sigma x\right)\sin\left(\beta x\right)}{x}\space\text{d}x\tag1$$
It is not hard to see that this is 'simple' Laplace transform:
$$\mathcal{I}\left(\sigma\space;\beta\right)=\mathscr{L}_x\left[\frac{\sin\left(\beta x\right)}{x}\right]_{\left(\sigma\right)}\tag2$$
Using the table of selected Laplace transforms and properties of the Laplace transform, we can see:
$$\mathcal{I}\left(\sigma\space;\beta\right)=\mathscr{L}_x\left[\frac{\sin\left(\beta x\right)}{x}\right]_{\left(\sigma\right)}=\int_\sigma^\infty\mathscr{L}_x\left[\sin\left(\beta x\right)\right]_{\left(\epsilon\right)}\space\text{d}\epsilon=$$
$$\int_\sigma^\infty\frac{\beta}{\epsilon^2+\beta^2}\space\text{d}\epsilon=\frac{1}{\beta}\int_\sigma^\infty\frac{1}{1+\left(\frac{x}{\beta}\right)^2}\space\text{d}\epsilon\tag3$$
Now, use a subsitution $\text{u}=\frac{x}{\beta}$ and your conclusion will follow.
A: After trying thousand finally solved!
I had found the RHS. So, I am taking that's equal to I.
$$I=\int_0^\infty \frac{e^{-ax}}{x}x\cos bx\mathrm dx$$
$$=\frac{a}{b}\int_0^\infty e^{-ax}\sin bx\mathrm dx$$
Since, $\sin\infty=0$ and, $\sin0=0$
If I integrate by parts again. Then, my equation looks like $$I=\frac{a}{b}-\frac{a^2}{b^2}I$$
So, $$I(1+\frac{a^2}{b^2})=\frac{a}{b}$$
Then, simple math.
