1/(x^2 + a^2)^2
with respect to x. At first sight this does not seem to admit an elementary solution. Indeed it is often used in examples of residue calculus to demonstrate the power of the method in calculating real integrals of rational functions! However, we note two other methods of dealing with it-
1) We employ the substitution x=a.tan u, whence the integral becomes one from -pi/2 to pi/2 of cos^2 (u) with respect to u. Using 2.cos^2 (u) = cos (2u) + 1 we then have the value to be pi/(2.a^3).
2) Note that the integral over the real line of,
1/(x^2 + a^2) = (1/a).[arctan (x/a)]_{x=-infinity}^{x=infinity} = pi/a
We now differentiate this relation with respect to a, effectively this is now a parameter integral, whence the integral is again pi/(2.a^3)
Posts
.jpg)