Feynman's Trick

A trick using the Leibniz Integral Rule to solve difficult Integrals

Example

$$\text{solve: }{\int }_0^{\infty }{e}^{-x^2}\cos (5x)dx$$ $$\begin{array}{c}\text{Let }f(\alpha )={\int }_0^{\infty }{e}^{-x^2}\cos (\alpha x)dx\\ \frac{d}{d\alpha }f(\alpha )=\frac{d}{d\alpha }{\int }_0^{\infty }{e}^{-x^2}\cos (\alpha x)dx\\ f^{\prime }(\alpha )={\int }_0^{\infty }{e}^{-x^2}\frac{\partial }{\partial \alpha }\cos (\alpha x)dx\\ f^{\prime }(\alpha )={\int }_0^{\infty }-x{e}^{-x^2}\sin (\alpha x)dx\\ \\ \begin{array}{cc}D& I\\ \sin (\alpha x)& -x{e}^{-x^2}\\ \alpha \cos (\alpha x)& \frac{1}{2}{e}^{-x^2}\end{array}\\ \\ f^{\prime }(\alpha )=\frac{1}{2}{e}^{-x^2}\sin (\alpha x){|}_0^{\infty }-\int \frac{1}{2}{e}^{-x^2}\alpha \cos (\alpha x)dx\\ f^{\prime }(\alpha )=0-\frac{\alpha }{2}f(\alpha )\\ f^{\prime }(\alpha )=-\frac{\alpha }{2}f(\alpha )\\ \frac{f^{\prime }(\alpha )}{f(\alpha )}=-\frac{\alpha }{2}\\ \int \frac{f^{\prime }(\alpha )}{f(\alpha )}d\alpha =\int -\frac{\alpha }{2}d\alpha \\ \int \frac{df}{f(\alpha )}d\alpha =-{\alpha }^2/4+C\\ \ln |f(\alpha )|=-{\alpha }^2/4+C\\ f(\alpha )=C{e}^{-{\alpha }^2/4}\\ \text{Let }\alpha =0:C=f(0)={\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}\\ f(\alpha )=\frac{\sqrt{\pi }}{2}{e}^{-{\alpha }^2/4}\\ \therefore f(5)={\int }_0^{\infty }{e}^{-x^2}\cos (5x)dx=\frac{\sqrt{\pi }}{2}{e}^{-25/4}\end{array}$$