Integration related to gamma function using rectangle contour

[Ex 9] Watson’s complex integration
$$\int^{\infty}_{-\infty}e^{-x^2}\,\cos(2ax)dx=e^{-a^2}\sqrt{\pi}$$
$$proof$$
Integrate the following function

$$f(z) = e^{-z^2}$$

Use the following contour

Note that the function is entire, hence
$$\int^{L}_{-L} e^{-t^2}\,dt+\int^{L+ai}_{L} e^{-t^2}\,dt+\int^{-L}_{-L+ai} e^{-t^2}\,dt+\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=0$$
For the forth integral use the substitution $x= t-ai$

$$\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=\int^{-L}_{L}e^{-(x+ai)^2}\,dx=-e^{a^2}\int^{L}_{-L}e^{-x^2}\,e^{-2iax}dx$$

Take $L \to \infty$

$$\left|\int^{L+ai}_{L} e^{-z^2}\,dz \right|=\left|\int^{ai}_{0} e^{-(x-L)^2}\,dt \right|\leq a e^{-L^2}\sim_{\infty} 0$$

Similarly

$$\left|\int^{-L}_{-L+ai} e^{-z^2}\,dz \right|=\left|\int^{ai}_{0} e^{-(x+L)^2}\,dt \right|\leq a e^{-L^2}\sim_{\infty} 0$$

Hence we dedeuce that

$$e^{a^2}\int^{\infty}_{-\infty}e^{-x^2}\,e^{-2iax}dx=\int^{\infty}_{-\infty} e^{-t^2}\,dt =\Gamma\left( \frac{1}{2}\right)=\sqrt{\pi}$$

By taking the real part and imaginary part

$$\boxed{\int^{\infty}_{-\infty}e^{-x^2}\,\cos(2ax)dx=e^{-a^2}\sqrt{\pi}}$$

This entry was posted in Contour Integration, Gamma function and tagged , , , , , , , . Bookmark the permalink.