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I know this integral evaluates to 1 by using numerical techniques. Can it be done analytically ?

$$ \int_{-\frac{1}{2}}^\infty \frac{2}{\sqrt{2y+1}\sqrt{2 \pi}} \exp \left( -y-\frac{1}{2} \right) dy $$

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Try substituting $y+\frac12 = t^2$. – mrf Oct 13 '12 at 22:45
up vote 3 down vote accepted

$$ u=y+\frac12, \qquad du=dy,\qquad 2y+1=2u $$ $$ \frac{2}{\sqrt{2\pi}}\int_{-1/2}^\infty \frac{1}{\sqrt{2y+1}} e^{-(y+(1/2))} \, dy = \frac{2}{\sqrt{2\pi}}\int_0^\infty \frac{1}{\sqrt{2u}} e^{-u}\,du = \frac{1}{\sqrt{\pi}}\int_0^\infty u^{-1/2} e^{-u}\,du. $$

The Gamma function is $$ \Gamma(\alpha)=\int_0^\infty u^{\alpha-1} e^{-u} \, du $$ and its value at $\alpha=1/2$ is the integral above. It is well known that $\Gamma(1/2)=\sqrt{\pi}$. I won't be surprised if the way that is proved is already an existing answer on stackexchange. If not, maybe we should put it here.

The value you seek is therefore $1$.

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Very helpful. +1. Thanks ! – Robert Long Oct 14 '12 at 6:37

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