8 Improper Integrals Using Contour Integration

We intend to evaluate convergent improper integrals of the form \[\int ^{\infty }_{-\infty } f(x)dx\quad \text {and}\quad \int ^{\infty }_0 g(x)dx\] provided the function \(g(x)\) can be restricted to a suitable complex function \(f(x)\).

This involves integrating a complex function \(f(z)\) around a closed contour \(\Gamma _R\) bounding a domain \(D\) of the type

The integral is evaluated using either the Cauchy integral theorem or the Cauchy Goursat, letting \(R\longrightarrow \infty ,\, r\longrightarrow 0\) we have the improper integrals.

If fig (a), the contour \(\Gamma _R\) to be traversed is the line \(OA\) along the positive \(x-\)axis, the circular arc \(C_R\) is of radius \(R\) centred at \(z = 0\) ate the vertical line \(BO\).

In fig (b), the contour \(\Gamma _R\) to be traversed is the line \(AB\) along the positive \(x-\)axis the circular arc \(C_R\) from \(B\) to \(C\) centred at \(z = 0\) with radius \(R\), the \(C_r\) from \(D\) to \(A\) with radius \(r\) centred at \(z = a\).

The construction in (b) is used when the function has a singularity at \(z = a\). This is called indenting the contour.

The value of the integral of \(f(z)\) around \(\Gamma _R\) determined either by Cauchy Goursat or \(C-I\) Cauchy integral is equal to the sum of the segments of \(\Gamma _R\).

For suitable \(f(z)\) the value of the integral along \(C_R\) tends to zero as \(R\longrightarrow \infty \) and if the contour is indented at the origin, the integral along \(C_r\) as \(r\longrightarrow 0\) can be found straight away.

A fundamental definite integral \[\int ^{\infty }_{-\infty } e^{-1/2\Big (\frac {x}{\rho }\Big )^2} dx = \rho \sqrt {2\pi }\qquad \Big (\text {Error function}\Big )\] from which we get \[\int ^{\infty }_0e^{-x^2}dx = \frac {1}{2}\sqrt {\pi }\]

In preparation to what comes next, we need to prove a simple but useful result - The Jordan inequality.

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