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I would like to compute the integral

$$ \int_{0}^{\infty}\frac{1}{\sqrt{2t}}e^{-\frac{1}{2t}}dt $$

which wolfram alpha says that it does not converge. However by letting $x=1/2t$ I get $dt=\frac{-1}{2x^2}dx$ and that the integral is transformed to $$ \int_{\infty}^{0}x^{0.5}e^{-x}\cdot\frac{-1}{2x^{2}}dx $$ or $$ \frac{1}{2}\int_{0}^{\infty}x^{-1.5}e^{-x}dx=0.5\Gamma(-0.5) $$ which is finite. Something must have gone wrong in the above calculation but I couldn't find out. Any help is appreciated.

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up vote 2 down vote accepted

The Gamma function is defined for complex $z$ with positive real part as $$ \Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt. $$ This integral does not converge if $\operatorname{Re}{z}\le0$. $\Gamma$ can be extended to the whole complex plane with the non-positive integers removed, but the above integral makes sense only if $\operatorname{Re}{z}>0$.

A similar situation happens for instance with power series. The series $\sum_{n=0}^\infty z^n$ converges on the unit disc to $1/(1-z)$. This function is defined on the whole plane wth the point $1$ removed, but agrees with the series only on the unit disc.

WolphramAlpha is right in telling that your integral does not converge. As $t\to\infty$, the integrand is like $1/\sqrt{2\,t}$.

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Thanks for the reply.. I've overlooked the definition that for negative $z$'s it cannot be written as the integral form. How can I rigorously prove that the integral does not converge? – David L Nov 26 '11 at 9:18
@David The integral is greater than $\int_0^{1}\frac{1}{\sqrt{2t}}\textrm{e}^{-\frac{1}{2t}}\ dt+\textrm{e}^{-1/2}\int_1^{\infty}\frac{1}{\sqrt{2t}}\ dt$. – alex.jordan Nov 26 '11 at 9:49
Thanks for the decomposition Alex. – David L Nov 26 '11 at 12:02

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