$\int^\infty_0 e^{-\alpha x}\sin(\beta x)\,dx = \frac{B}{\alpha^2+\beta^2}$ Laplace

Question

$$ \int^\infty_0 \! e^{-\alpha x} \sin(\beta x)\,dx = \frac{\beta}{\alpha^2+\beta^2} $$

Can someone start this for me? I don't know where to begin.

Answer 1

Maybe the easiest (or at least most economic) way to solve this is using the fact that $\sin(\beta x)=\Im[{e^{i\beta x}}]$:

$$ I=\Im\left[\int_0^{\infty}{e^{i\beta x- ax}}\right]=-\Im\left[\frac{1}{i\beta-a}\right]=\frac{\beta}{\beta^2+a^2} $$

Answer 2

One approach is to integrate by parts twice $$ \begin{align} &\int_0^\infty e^{-\alpha x}\sin(\beta x)\,\mathrm{d}x\tag{1}\\ &=-\frac1\alpha\int_0^\infty\sin(\beta x)\,\mathrm{d}e^{-\alpha x}\tag{2}\\ &=\frac\beta\alpha\int_0^\infty e^{-\alpha x}\cos(\beta x)\,\mathrm{d}x\tag{3}\\ &=-\frac\beta{\alpha^2}\int_0^\infty\cos(\beta x)\,\mathrm{d}e^{-\alpha x}\tag{4}\\ &=\frac\beta{\alpha^2}-\frac{\beta^2}{\alpha^2}\int_0^\infty e^{-\alpha x}\sin(\beta x)\,\mathrm{d}x\tag{5}\\[3pt] &=\frac\beta{\alpha^2+\beta^2}\tag{6} \end{align} $$ Explanation:
$(2)$: $e^{-\alpha x}\,\mathrm{d}x=-\frac1\alpha\,\mathrm{d}e^{-\alpha x}$
$(3)$: integrate by parts
$(4)$: $e^{-\alpha x}\,\mathrm{d}x=-\frac1\alpha\,\mathrm{d}e^{-\alpha x}$
$(5)$: integrate by parts
$(6)$: add $\frac{\beta^2}{\alpha^2+\beta^2}$ times $(1)$ to $\frac{\alpha^2}{\alpha^2+\beta^2}$ times $(5)$