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Improper integral (ThinleyDs question at Yahoo! Answers)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello ThinleyD,

Necessarily $b>0$, otherwise the integral is divergent. Using the substitution $t=\sqrt{b}(x-a)$: $$\int_{-\infty}^{+\infty}e^{-b(x-a)^2}\;dx=\frac{1}{\sqrt{b}}\int_{-\infty}^{+\infty}e^{-t^2}\;dt=\frac{2}{\sqrt{b}}\int_{0}^{+\infty}e^{-t^2}\;dt$$ We get the well known Euler's integral. Using $u=t^2$: $$\int_{0}^{+\infty}e^{-t^2}\;dt=\int_{0}^{ + \infty} e^{-u}\frac{du}{2\sqrt{u}}=\frac{1}{2}\int_{0}^{ +\infty}e^{-u}u^{-\frac{1}{2}}\;du=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$$ As a consequence: $$\boxed{\displaystyle\int_{-\infty}^{+\infty}e^{-b(x-a)^2}\;dx=\sqrt{\frac{\pi}{b}}}$$