Can complex analysis be used to evaluate real definite integrals?

[zetafunction]

given a function g(x) so the integral \int_{-\infty}^{\infty}dx g(x) exists

could we find a complex valued function f(z) and a closed curve C so

\int_{-\infty}^{\infty}dx g(x)= \oint _ {C} dz f(z)

then if we can calculate the residues of f(z) we can compute the real valued integral of g(x)

is this possible for any well-behaved functions f and g(x) ??

 

[HallsofIvy]

No, but what you could do, for example, is define f(z) such that is z= x+ 0i (i.e. if z is real) then f(z)= g(x). Then take the integral over, say, the contour formed by the straight line from (-R, 0) to (R, 0) along the real axis, then along the circle of radius R, with center at (0,0), from (R, 0) to (-R, 0). That will be, of course, \int_{-R}^R g(x)dx+ \int f(z)dz where the second integral is over the semi-circle and will be equal to the sum of the residues of f(z) inside the semicircle. Taking the limit as R goes to infinity, if the integral around the half circle goes to 0, you have \int_{-\infy}^\infty f(x) dx equal to the sum of the residues in the upper half plane.

 

[g_edgar]

The topic of evaluating real definite integrals using contour integrals should be found in textbooks in complex analysis. It is something of an art to find the right function and the right contour. And usually there is a limit at the end, too. I saw a fairly recent (as such things go) paper where somebody figured out how to get
\int_{-\infty}^{+\infty} e^{-x^2}\,dx
by this method, but the trick was to take a paralellogram as the contour.