Evaluate the following integral using Laplace transform $$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$

I obtained this partial result $$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$ and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer.

  • $\begingroup$ Since you know the answer depends on $a$, when you arrive at an intermediate stage which doesn't mention $a$, you know you must have gone wrong... $\endgroup$ – AakashM Jul 5 '17 at 12:38
  • $\begingroup$ If you're still confused remember that $\tan^{-1}(s)+\tan^{-1}(1/s)=\pi/2$ $\endgroup$ – user5389726598465 Jul 5 '17 at 21:25

Your partial result does not hold. According to Laplace transform properties $$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx=\cal{L}\left(\frac{\sin bx}{x}\right)(a)=\int_a^{+\infty} \cal{L}\left(\sin bx\right)(p)dp=\int_a^{+\infty} \frac{b}{p^2+b^2}\, dp.$$ Can you take it form here?


the standard form of LT is $$\mathcal{L}f(t)=\int_{0}^{\infty}e^{-st}f(t)\,dt$$ So for the given integral you just have to calculate LT of $\frac{\sin bx}{x}$ then put $s=a$ these two properties of LT may be useful $$\mathcal{L} \sin (bx)=\frac{b}{b^2+s^2}$$and $$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_{s}^{\infty}F(s)\,ds$$


This doesn't use the Laplace Transform, however, it is pretty simple.

$$ \begin{align} \frac{\partial}{\partial a}\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x &=-\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}x\tag{1}\\ &=-\frac1b+\frac ab\int_0^\infty e^{-ax}\cos(bx)\,\mathrm{d}x\tag{2}\\ &=-\frac1b+\frac{a^2}{b^2}\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}\tag{3}\\ &=-\frac{b}{a^2+b^2}\tag{4} \end{align} $$ Explanation:
$(1)$: differentiate inside the integral with respect to $a$
$(2)$: integrate by parts: $-e^{-ax}\sin(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\cos(bx)$
$(3)$: integrate by parts: $e^{-ax}\cos(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\sin(bx)$
$(4)$: $\frac{b^2}{a^2+b^2}$ times $(3)$ plus $\frac{a^2}{a^2+b^2}$ times $(1)$

Integrating $(4)$ with respect to $a$ gives $$ \int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x=\tan^{-1}\left(\frac ba\right)\tag{5} $$

  • 3
    $\begingroup$ Feynmann to the rescue !! $\endgroup$ – Siddhartha Ganguly Jul 5 '17 at 17:04

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