Improper complex integration

Question

I was trying the problem of Spiegel complex variables chapter 4 prob 93 :

$$\int_{0}^{\infty}xe^{-x}\sin x\, \mathrm dx = \frac12$$

I tried with by parts and and put the limits... but the ans is not the same... Please let me know whether there is an easy way to do this without by parts... Thank you.

Answer 1

A tricky attack: since $$\mathcal{L}(\sin x)=\frac{1}{s^2+1},\tag{1}$$ we have: $$\mathcal{L}(x\cdot\sin x) = -\frac{d}{ds}\mathcal{L}(\sin x) = \frac{2s}{(s^2+1)^2}\tag{2}$$ and by evaluating the RHS of $(2)$ at $s=1$ we get: $$ \int_{0}^{+\infty} xe^{-x}\sin x\,dx = \frac{2\cdot 1}{(1+1)^2}=\color{red}{\frac{1}{2}}\tag{3}$$ as wanted.

Answer 2

Define our Integral as $J$ ,furthermore let's set

$$ I[a]=\int_{0}^{\infty}e^{-ax+ix} $$

Then $$ J=-\Im\left(\partial_aI[a]\right)\big|_{a=1}=\Im\left(\partial_a\left(\frac{1}{-a+i}\right)\right)\big|_{a=1} $$

I'm sure you can take it from here!