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A.5 Example 2: Complex exponentials

The most common problem is with complex exponentials (or sines and cosines which can be written as such).

Calculate the integral (which falls slowly for large x!)

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞} {{e}^{ikx}\over i − x}\kern 1.66702pt dx
(A.14)

We shall analyse this for k > 0.

If we substitute x = z = R{e}^{iϕ}, we find

The problem can be seen on substitution of z = R{e}^{iϕ}, for R fixed (as a bove)

{{e}^{ikx}\over i − x} = {{e}^{−kR\mathop{ sin}\nolimits (ϕ)}(\mathop{cos}\nolimits (kR\mathop{cos}\nolimits (ϕ)) + i\mathop{sin}\nolimits (kR\mathop{cos}\nolimits (ϕ)))\over −R\mathop{cos}\nolimits (ϕ) − iR\mathop{sin}\nolimits (ϕ) + i}

For \mathop{sin}\nolimits (ϕ) > 0 the integrand goes to zero very quickly with R, but for ϕ=0 we enter a grey territory, where the integrand decays like 1∕R. If we move the original integral up by just a little bit (ϵ) we are OK, since ϕ doesn’t become zero. Thus

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞} {{e}^{ikx}\over i − x}\kern 1.66702pt dx ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞+iϵ}^{∞+iϵ} {{e}^{ikz}\over i − z}\kern 1.66702pt dz = \mathop{∮ }\nolimits {{e}^{ikz}\over i − z}dz
(A.15)

The residue is easily seen to be R = −{e}^{−k}, and thus

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞} {{e}^{ikx}\over i − x}\kern 1.66702pt dx ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞+iϵ}^{∞+iϵ} {{e}^{ikz}\over i − z}\kern 1.66702pt dz = \mathop{∮ }\nolimits {{e}^{ikz}\over i − z}dz = 2πi\left (−{e}^{−k}\right ) = −2πi\text{ }{e}^{−k}\text{ }(k > 0)
(A.16)

In the same way we can show that for k < 0 we must close the contour in the lower plane (since k\mathop{sin}\nolimits (ϕ) must be negative)

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞} {{e}^{ikx}\over i − x}\kern 1.66702pt dx ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞−iϵ}^{∞−iϵ} {{e}^{ikz}\over i − z}\kern 1.66702pt dz = \mathop{∮ }\nolimits {{e}^{ikz}\over i − z}dz = −2πi(0) = 0\text{ }(k < 0)
(A.17)

since no pole is enclosed inside the contour.

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