Question:

Find the value of the following integral:

\[  \int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx  \]

Answer:

Let

\[  \int^{\infty}_{-\infty}f(x,b)dx = 2\int^{\infty}_{0}f(x,b)dx  \]

where \(f(x,b)=e^{-x^2}\cos 2bx\). According to the properties of the functions, we can state that \(|f(x,b)| \leq e^{-x^2}\); and \(\int^{\infty}_{0}e^{-x^2}\) is finite. Therefore, \(\varphi(b)=\int^{\infty}_{0}f(x,b)dx\) is also finite. Now, take the partial derivative of the function \(f(x,b)\) with respect to \(b\).

\[  f_b(x,b) = -2xe^{-x^2}\sin 2bx  \]

This is continuous for the arbitrary value of \(b\) with \(x \geq 0\). Also we know \(|f(x,b)| \leq 2xe^{-x^2}\); and \(\int^{\infty}_{0}2xe^{-x^2}\) is finite. Therefore, \(\int^{\infty}_{0}f_b(x,b)dx\) is also finite for any \(b\). Then,

\[  \varphi'(b) = \int^{\infty}_{0}f_b(x,b)dx  \]

\[  =  \int^{\infty}_{0}-2xe^{-x^2}\sin 2bxdx  \]

\[   = \left[ e^{-x^2}\sin 2bx \right]^{\infty}_{0} – 2b\int^{\infty}_{0}e^{-x^2}\cos 2bxdx \]

\[   = 0-2b\varphi(b)  \]

From the above, we can have the following relationship:

\[  \frac{\varphi'(b)}{\varphi(b)}=-2b  \]

Then, integrate both sides.

\[  \ln|\varphi(b)| = -b^2 + C  \]

\[  \varphi(b) = Ce^{-b^2}  \]

From the initial condition,

\[  \varphi(0) = \int^{\infty}_{0}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}=C  \]

Thus,

\[  \varphi(b)=\frac{\sqrt{\pi}}{2}e^{-b^2}  \]

Therefore,

\[  \int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx = \sqrt{\pi}e^{-b^2}  \]