1. ## difficult improper integral

integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.

2. Originally Posted by oblixps
integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.

Its an unelementary integral.
Post the whole problem. The problem asks you to find its value or just determine its convergence/divergence?

3. Originally Posted by oblixps
integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.
You have it. First recall $\displaystyle \int_0^{\infty} e^{-w^2}dw=\frac{\sqrt{\pi}}{2}$

Now you have $\displaystyle 2\int_0^{\infty}e^{-2u^2}du$. How about letting $\displaystyle w=\sqrt{2}u$

4. Originally Posted by shawsend
You have it. First recall $\displaystyle \int_0^{\infty} e^{-w^2}dw=\frac{\sqrt{\pi}}{2}$

Now you have $\displaystyle 2\int_0^{\infty}e^{-2u^2}du$. How about letting $\displaystyle w=\sqrt{2}u$
There is a well known trick to evaluating this last integral, which is to consider:

$\displaystyle I^2=\left[\int_{-\infty}^{\infty}e^{-u^2}du\right]^2=\int_{-\infty}^{\infty}e^{-x^2}dx \int_{-\infty}^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy$

Now we can rewrite the last integral in polars instead of cartesians:

$\displaystyle \int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy=\int_{r=0}^{\infty}\int_{\theta=0}^ {2\pi}e^{-r^2} rd\theta dr$

which can be evaluated by elementary means.

CB

5. Originally Posted by CaptainBlack
There is a well known trick to evaluating this last integral, which is to consider:

$\displaystyle I^2=\left[\int_0^{\infty}e^{-u^2}du\right]^2=\int_0^{\infty}e^{-x^2}dx \int_0^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy$

Now we can rewrite the last integral in polars instead of cartesians:

$\displaystyle \int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy=\int_{r=0}^{\infty}\int_{\theta=0}^ {2\pi}e^{-r^2} rd\theta dr$

which can be evaluated by elementary means.

CB
would the limits on theta be from 0 to pi/2? since the original limits of e^(-x^2) are from 0 to infinite so we are considering the first quadrant.

6. Originally Posted by oblixps
would the limits on theta be from 0 to pi/2? since the original limits of e^(-x^2) are from 0 to infinite so we are considering the first quadrant.
There is a mistake in my posting the first two integrals should be from $\displaystyle -\infty$ to $\displaystyle +\infty$, then I want to leave it to the reader to divide $\displaystyle I$ by $\displaystyle 2$ to get what you asked for. But changing the integral to be the first quadrant works as well (it shift where we divide by 2 to a different place but just as valid).

CB