# 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!

• Try to differentiate the integral with respect to $a$, the resulting integrand has elementary antiderivative. – pisco Aug 9 '15 at 7:24

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

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)}}$$

Hint: the integral is definition for Laplace transformation of sin(bx) / x

$\bf{My\; Solution::}$ 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 $$\displaystyle I= \int_{0}^{b}\int_{0}^{\infty}e^{-ax}\cdot \cos(yx)dxdy = \int_{0}^{b}\frac{a}{y^2+a^2}dy = \tan^{-1}\left(\frac{b}{a}\right)$$

• I think there would be $$\displaystyle \int e^{-ax}\cdot \sin bx = -\frac{e^{-ax}}{a^2+b^2}\left(a\sin (bx)+b\cdot \cos (bx)\right)$$ – Pratyush Aug 9 '15 at 7:43
• This solution is okay for computing $I(a,b)$ when $a > 0$, but the problem is more subtle to compute $I(0,b)$ since the integral is not absolutely convergent when $a = 0$. – KCd Aug 9 '15 at 7:52