# How to show that for $|a|<1$: $\int_0^\infty\frac{x^a}{(x+2)^2}\,dx=\frac{a\pi2^{a-1}}{\sin\pi a}$?

Show that for $|a|<1$: $$\int_0^\infty\frac{x^a}{(x+2)^2}\,dx=\frac{a\pi2^{a-1}}{\sin\pi a}$$

• $x = 2 y$ gives $2^{a-1}\int_0^\infty \frac{y^a}{(y+1)^2} d y$ – Gribouillis Dec 11 '17 at 15:57
• What have you attempted? Have you studied contour integration, differentiation under the integral sign or the $\Gamma$ function? – Jack D'Aurizio Dec 11 '17 at 16:22

Through the Laplace transform $$\mathcal{L}(x^a) = \frac{\Gamma(a+1)}{s^{a+1}},\qquad \mathcal{L}^{-1}\left(\frac{1}{(x+2)^2}\right) = s e^{-2s}$$ hence the original integral equals $$\Gamma(a+1)\int_{0}^{+\infty} s^{-a} e^{-2s}\,ds = 2^{a-1}\Gamma(1+a)\Gamma(1-a)= a 2^{a-1} \Gamma(a)\Gamma(1-a) = \frac{\pi a 2^{a-1}}{\sin(\pi a)}$$ by the reflection formula for the $\Gamma$ function.
With $\ds{\verts{\Re\pars{a} < 1}}$: