# Complex analysis and Cauchy Residue theorem

There is a small formula for finding definite integral by complex analysis method : If $$f(x)$$ containes cosine and sine functions along with polynomial functions then $$f(x)$$ can be treated as a real or imaginary part of $$f(z)$$.Then find the singular points of $$f(z)$$ and check which point lies in upper half plane.

CASE(a):If the singular point does not lie on real axis in this case we apply Cauchy residue theorem and may use Jordan's lemma and everything is claear to me .

CASE(b):If the singular points lie on real axis ,then:

      Method 1: by using cauchy residue thm and taking a semicircular contour which leaves the points of singularity on real axis.

Method 2:(short one)


$$\int_{-\infty}^\infty f(z)dz$$ = $$\pi i$$ [$$\sum$$ residue at poles within C] ,if jordan's lemma applied.

now my question is what is proof for this short method .I thought of many things for why is there a factor of $$\pi i$$ in place of $$2\pi i$$ but they don't look right.

• You avoid the singularity by taking a semicircular contour and shrinking that to zero. You find that the fact a semicircle subtends an angle of $\pi$ is significant. Jul 29, 2020 at 14:56
• Yes it is, but it that way we are finding residue without taking a closed contour around singularity. Jul 29, 2020 at 15:04
• You are not evaluating the residue, but using Jordan's lemmata instead. Jul 29, 2020 at 15:26

The answer will be $$\pm i \pi\, \text{Residuum}$$ depending on how you "regularize" the singularity -- by pushing it up or down away from the real axis. It is equivalent to integration over a half-circle above or under the singularity and shrinking the radius to zero. The factor $$i \pi$$ comes from the half-circle integration exactly as the factor $$2 \pi i$$ comes from the full-circle integration.

There is also a prescrption called principal value in which one effectively takes half of the upper and half of the lower half-circle values. It depends on the problem at hand which prescription should be used.

Of course, assuming also that you can close the integration contour by a semicircle at infinity.