Integrate $I=\int_0^\infty \frac{e^{-ax}\sin bx}{x}\mathrm dx$

Question

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.)

$\int_0^{\infty} e^{-ax} \sin (bx)dx$ and $\int_0^{\infty} e^{-ax} \cos (bx)dx$ are standard integrals. They are evaluated using two integrations by parts and you can find these in most Calculus books. — Kavi Rama Murthy, Aug 03 '21 at 05:13
@KaviRamaMurthy Sorry! I didn't give the whole question. And, I am doing improper integrals not standard one. — , Aug 03 '21 at 05:14
Using Laplace Transformation $L{\sin bx } =\frac b{s^2+b^2}$ Hence $L{\frac {\sin bx} {x} } =\int_s^{\infty} \frac b{u ^2+b^2} du ={\tan ^{-1} \frac {u} b}s^{\infty}=\cot ^{-1}\frac sb$ Hence $L{\frac {e^{ax} \sin bx}{x} } ={\cot ^{-1}\frac sb}{s=a} =\cot ^{-1} \frac ab =\tan ^{-1} \frac ba $ — Math-Learner, Aug 03 '21 at 06:38
@Istiak You can find your answer right here: https://math.stackexchange.com/questions/3439419/evaluating-int-0-infty-e-ax-frac-sinbxxdx?rq=1 — Arjun Vyavaharkar, Aug 03 '21 at 06:44
Does this answer your question? Evaluating $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx $ — Arjun Vyavaharkar, Aug 03 '21 at 06:46
You keep doing 50% of the work and stop right there. MSE is not the site to be spoonfed. (-1) — Arjun Vyavaharkar, Aug 03 '21 at 06:49

score 1 · Answer 1 · answered Aug 03 '21 at 13:07

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.

score 0 · Accepted Answer · 2021-08-03T08:37:31.543

0

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.

edited Aug 03 '21 at 08:37

answered Aug 03 '21 at 08:03

The question asked itself is incorrect. It should be $e^{-ax}$ . – Laxmi Narayan Bhandari Aug 03 '21 at 08:14
@LaxmiNarayan ohh! Sorry that was a typo – Aug 03 '21 at 08:37

Integrate $I=\int_0^\infty \frac{e^{-ax}\sin bx}{x}\mathrm dx$

2 Answers2