integral from 0 to infinity of 1/(pi^2+x^2)^(3/2)dx

Answer 1

if x = pi*tan(z)
pi^2 + x^2 = pi^2 + pi^2 tan^2(z) = pi^2 sec^2(z)
dx = pi*sec^2(z)

1/(pi^3 sec^3(z)) * pi*sec^2(z) dz
= 1/pi^2 cos(z) dz

the integral is thus 1/pi^2 sin(z)

x/pi * tan(z), so
sin(z) = x/sqrt(pi^2 + x^2)

the integral is thus x/[pi^2 sqrt(pi^2 + x^2)]

from 0 to infinity is thus 1/pi^2