How do you find the integral of #int x/((x^2 +2)^2) dx# from 0 to infinity?

Question

Elijah Abram · Accepted Answer

#int_0^oo x/(x^2+2)^2dx = 1/2int_0^oo(2x)/(x^2+2)^2dx# Let's #u = x^2+2# So #du = 2xdx# #1/2int_2^oo1/(u^2)du# #-1/2[1/u]_2^oo# #=-1/2(1/oo-1/2) = -1/2(0-1/2)#
#=1/4#

Paisley Johnson · Answer

undefined To find the integral of $ \int_0^\infty \frac{x}{(x^2 + 2)^2} \, dx $, you can use the method of complex integration.

1. Let $ f(z) = \frac{z}{(z^2 + 2)^2} $.
2. Consider the contour integral $ \oint_C f(z) \, dz $, where $ C $ is a closed contour consisting of a semicircle in the upper half-plane from $ -R $ to $ R $, and a straight line segment from $ R $ to $ -R $.
3. Apply Cauchy's residue theorem, which states that the contour integral is equal to $ 2\pi i $ times the sum of residues of $ f(z) $ at its singular points within $ C $.
4. Find the residues of $ f(z) $ at its singular points, which are the poles of $ f(z) $ within the contour.
5. Evaluate the integral by taking the limit as $ R $ approaches infinity and applying the residue theorem.

After performing these steps, you will find that the integral equals $ \frac{\pi}{\sqrt{2}} $.