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 {\bf Question}

 Assume that $\ds\int_{-\infty}^{\infty}
\frac{1}{\sqrt{2\pi b}}e^{-\frac{1}{2b^2}(y-a)^2} \,dy=1$ where
$b>0$.

Evaluate the integral $\ds\int_0^{\infty}  e^{-3y^2} \,dy$.


\medskip

 {\bf Answer}

We know that

$\ds \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}b} e^{-
\frac{(y-a)^2}{2b^2}} \,dy = 1$

Therefore $\ds \int_{-\infty}^{\infty}  \frac{1}{\sqrt{2\pi}b}
e^{-\ds \frac{y^2}{2b^2}} \,dy = 1$

i.e. $\ds 2\int_0^{\infty} \frac{1}{\sqrt{2\pi}b} e^{-
\frac{y^2}{2b^2}} \,dy = 1$

i.e. $\ds \int_0^{\infty} e^{- \frac{y^2}{2b^2}} \,dy = \ds
\frac{\sqrt{2\pi}b}{2}$

Let $\ds \frac{1}{2b^2}  =  3 \Rightarrow  b^2 = \frac{1}{6}
\Rightarrow  b = \sqrt{\frac{1}{6}}$

Therefore $\ds \int_0^{\infty} e^{-3y^2} \,dy  =
\frac{\sqrt{2\pi}}{2} \sqrt{\frac{1}{6}} =
\frac{\sqrt{\pi}}{2\sqrt{3}}$


\end{document}