Evaluate:$I=\int^\infty_0 \frac {e^{-ax}\ \sin bx}{x}\,dx$ How would I evaluate the integral:
$$I=\int^\infty_0 \frac {e^{-ax}\ \sin bx}{x}\,dx$$
I have started doing the problem by integration by parts but it seems to be more lengthy. Since it is definite integration, any properties may be there. I can't figure out the property basically. Any suggestion will be highly appreciated. Thanks!
 A: Notice, using property of Laplace transform as follows 
$$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$$
$$L(\sin bt)=\int_{0}^{\infty}e^{-st}\sin t dt=\frac{b}{b^2+s^2}$$
Now, we have 
$$\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx$$
$$=\int_{a}^{\infty} L(\sin bx)dx$$
$$=\int_{a}^{\infty}\frac{b}{b^2+x^2} dx$$
$$=b\int_{a}^{\infty}\frac{dx}{b^2+x^2} $$
$$=b\left[\frac{1}{b}\tan^{-1}\left(\frac{x}{b}\right)\right]_{a}^{\infty} $$
$$=\left[\tan^{-1}\left(\infty\right)-\tan^{-1}\left(\frac{a}{b}\right)\right] $$ $$=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)$$ Hence, we have
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)}}$$
A: Hint: the integral is definition for Laplace transformation of sin(bx) / x
A: $\bf{My\; Solution::}$ Let $$\displaystyle I(a,b) = \int_{0}^{\infty}\frac{e^{-ax}\cdot \sin (bx)}{x}dx$$
Now Fiff. both side w. r to $a\;,$ We get
$$\displaystyle \frac{dI(a,b)}{da} = \frac{d}{da}\int_{0}^{\infty}\frac{e^{-ax}\cdot \sin (bx)}{x}dx = \int_{0}^{\infty}\frac{e^{-ax}\cdot -x\cdot \sin (bx)}{x}dx = -\int_{0}^{\infty}e^{-ax}\cdot \sin (bx)dx$$
Now $$\displaystyle \int e^{-ax}\cdot \sin bx = -\frac{e^{-ax}}{a^2+b^2}\left(a\cdot \sin (bx)+b\cdot \cos (bx)\right)$$
(Above we have used Integration By Parts)
So $$\displaystyle \int_{0}^{\infty}e^{-ax}\cdot \sin bx = -\frac{b}{a^2+b^2}$$
So We get $\displaystyle \frac{dI(a,b)}{da} = \frac{b}{a^2+b^2}\Rightarrow \displaystyle \int \frac{dI(a,b)}{da}da = \int\frac{b}{a^2+b^2}da$
$\underline{\bf{Another\; way::}}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{e^{-ax}\cdot \sin (bx)}{x}dx = \int_{0}^{\infty}\int_{0}^{b}e^{-ax}\cdot \cos(yx)dydx$$
So we write $$\displaystyle I= \int_{0}^{b}\int_{0}^{\infty}e^{-ax}\cdot \cos(yx)dxdy = \int_{0}^{b}\frac{a}{y^2+a^2}dy = \tan^{-1}\left(\frac{b}{a}\right)$$
