# Math Help - Integral

1. ## Integral

Will someone check my work....I think this is right

$\int_0^{\infty}e^{-a^2x^2}dx,\forall{a}>0$

First $Let\text{ }u=a^2x^2\Rightarrow{dx=\frac{1}{2a\sqrt{u}}}$

and $u(0)=a^2\cdot{0^2}=0$

and $u(\infty)=a^2\cdot{\infty}=\infty$

So this would be $\frac{1}{2a}\int_0^{\infty}\frac{e^{-u}}{\sqrt{u}}du=\frac{1}{2a}\int_0^{\infty}u^{\fra c{-1}{2}}e^{-u}du=\frac{\Gamma\bigg(\frac{1}{2}\bigg)}{2a}=\fra c{\sqrt{\pi}}{2a}$

This doesnt seem right...I mean I think I did my calculations right but this was in a very simple book and this seems uncharcteristic of such a book ....

Could someone verify?

2. Originally Posted by Mathstud28
Will someone check my work....I think this is right

$\int_0^{\infty}e^{-a^2x^2}dx,\forall{a}>0$

First $Let\text{ }u=a^2x^2\Rightarrow{dx=\frac{1}{2a\sqrt{u}}}$

and $u(0)=a^2\cdot{0^2}=0$

and $u(\infty)=a^2\cdot{\infty}=\infty$

So this would be $\frac{1}{2a}\int_0^{\infty}\frac{e^{-u}}{\sqrt{u}}du=\frac{1}{2a}\int_0^{\infty}u^{\fra c{-1}{2}}e^{-u}du=\frac{\Gamma\bigg(\frac{1}{2}\bigg)}{2a}=\fra c{\sqrt{\pi}}{2a}$

This doesnt seem right...I mean I think I did my calculations right but this was in a very simple book and this seems uncharcteristic of such a book ....

Could someone verify?
$\int_{0}^{\infty}e^{-a^2x^2}\,dx \ \forall \ a>0$
$let \ u=a^2x^2 \implies du=2a^2x\,dx\implies dx=\frac{du}{2a^2\sqrt{\frac{u}{a^2}}}\implies dx=\frac{du}{2a\sqrt{u}}$

$u(0)=0 \ and \ u(\infty)=\infty$.

$\therefore \frac{1}{2a}\int_{0}^{\infty}e^{-u}u^{-\frac{1}{2}}\,du=\frac{\Gamma\left(\frac{1}{2}\rig ht)}{2a}=\frac{\sqrt{\pi}}{2a}$.

I get what you get.

3. Originally Posted by Chris L T521
$\int_{0}^{\infty}e^{-a^2x^2}\,dx \ \forall \ a>0$
$let \ u=a^2x^2 \implies du=2a^2x\,dx\implies dx=\frac{du}{2a^2\sqrt{\frac{u}{a^2}}}\implies dx=\frac{du}{2a\sqrt{u}}$

$u(0)=0 \ and \ u(\infty)=\infty$.

$\therefore \frac{1}{2a}\int_{0}^{\infty}e^{-u}u^{-\frac{1}{2}}\,du=\frac{\Gamma\left(\frac{1}{2}\rig ht)}{2a}=\frac{\sqrt{\pi}}{2a}$...

I get what you get.
I didnt think this was right because this was juxtaposed beside this integral

$\int_2^3\frac{dx}{1-\cos(x)}$

So I was like...hmm