Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $1/\sqrt{2\,t}$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.