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NOTE REFERENCE TO QUESTION 10

{\bf Question}

Verify the following integral.

$\ds\frac{1}{2\pi i} \ds\int_{a-i\infty}^{a+i\infty} \,dz
\ds\frac{\exp(zt)}{\sqrt{z+1}}=\ds\frac{\exp(-t)}{\sqrt{\pi t}},\
a&gt;0$

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{\bf Answer}

$I=\ds\frac{1}{2\pi i}\ds\int_{a-i\infty}^{a+i\infty}
\ds\frac{e^{zt}}{\sqrt{z+1}}dz$ put $z+1=\zeta$ say

$\ \ =\ds\frac{1}{2\pi i}\ds\int_{a-1-i\infty}^{a+1+i\infty}
\ds\frac{e^{\zeta t}}{\sqrt{\zeta}}e^{-t}d\zeta=e^{-t} \times$
integral of Q10

Use same arguments as in Q10 and you get
$I=\ds\frac{e^{-t}}{\sqrt{\pi t}}$
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