# analytic integral

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

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