1. ## intergration help!!!

evalutate the intergral

intergral between infinity and 0 of $e^{-x^{2}}$

2. Originally Posted by turion645344
evalutate the intergral

intergral between infinity and 0 of $e^{-x^{2}}$

You mean $\int_0^{\infty} e^{-x^2}dx$ I presume?

There is a neat little trick to doing this.

First notice that the integrand is an even function. So
$\int_0^{\infty} e^{-x^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx$

Now I am going to integrate $\int_{-\infty}^{\infty} e^{-x^2} dx$ because it is simpler than the original.

Call the value of the integral
$I = \int_{-\infty}^{\infty} e^{-x^2} dx$.
Then
$I^2 = \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dy$
where x and y are independent variables.

$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy$

Now, define an xy plane. We are integrating over the whole of the xy plane. I am going to choose to do this with polar, rather than rectangular, coordinates. So define:
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$
in the usual manner.

Then $dx dy = r dr d\theta$ and $x^2 + y^2 = r^2$. So...
$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} r dr d\theta$

The $\theta$ integration is trivial:
$I^2 = 2\pi \cdot \int_0^{\infty} e^{-r^2} r dr$

Substitute $z = r^2$, then $dz = 2r dr$, so
$I^2 = 2\pi \cdot \frac{1}{2} \int_0^{\infty} e^{-z} dz$

$I^2 = \pi \cdot -e^{-z}|_0^{\infty} = -pi \cdot (0 - 1) = \pi$

Thus
$I = \sqrt{\pi} = \int_{-\infty}^{\infty} e^{-x^2}dx$

So finally:
$\int_0^{\infty} e^{-x^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$

-Dan

3. Originally Posted by topsquark
You mean $\int_0^{\infty} e^{-x^2}dx$ I presume?

There is a neat little trick to doing this.

First notice that the integrand is an even function. So
$\int_0^{\infty} e^{-x^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx$

Now I am going to integrate $\int_{-\infty}^{\infty} e^{-x^2} dx$ because it is simpler than the original.

Call the value of the integral
$I = \int_{-\infty}^{\infty} e^{-x^2} dx$.
Then
$I^2 = \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dy$
where x and y are independent variables.

$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy$

Now, define an xy plane. We are integrating over the whole of the xy plane. I am going to choose to do this with polar, rather than rectangular, coordinates. So define:
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$
in the usual manner.

Then $dx dy = r dr d\theta$ and $x^2 + y^2 = r^2$. So...
$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} r dr d\theta$

The $\theta$ integration is trivial:
$I^2 = 2\pi \cdot \int_0^{\infty} e^{-r^2} r dr$

Substitute $z = r^2$, then $dz = 2r dr$, so
$I^2 = 2\pi \cdot \frac{1}{2} \int_0^{\infty} e^{-z} dz$

$I^2 = \pi \cdot -e^{-z}|_0^{\infty} = -pi \cdot (0 - 1) = \pi$

Thus
$I = \sqrt{\pi} = \int_{-\infty}^{\infty} e^{-x^2}dx$

So finally:
$\int_0^{\infty} e^{-x^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$

-Dan
I've seen this trick a long, long time ago, but I forgot all about it. I would have been stumped on this integral

4. Originally Posted by Jhevon
I've seen this trick a long, long time ago, but I forgot all about it. I would have been stumped on this integral
I actually learned it in High School and it fascinated me so much that I remembered it.

Too bad I have only rarely been able to apply this trick to other integrals I've run into.

-Dan

5. I have two other approaches to this integral. One of them uses manipulation of the Beta and Gamma Integrals which I do not find to be the nicest approach. Here is another approach I present below, it assumes knowledge of Contour Integration.
-----
1)Define the function $f(z) = e^{i\pi z^2}$ and let $g(z) = \frac{f(z)}{\sin \pi z}$.

2)Contrust the paralleogram on $\pm Re^{i\pi /4} \pm \frac{1}{2}$ for a big positive number $R$ (which we will sent to $+\infty$).

3)Look at picture below, just be warned I did not write the $\pi i/4$ exponent of $e$ because it is difficult through MS-Paint. The red dot is the only pole for $g(z)$ which occurs as $z=0$.

4) $\lim_{z\to 0}z\cdot \frac{f(z)}{\sin \pi z} = \frac{1}{\pi}$ so $\mbox{res}(g,0) = \frac{1}{\pi}$.

5)The right sloped sides can be parametrized as $\gamma_1(t) = e^{\pi i/4}t+\frac{1}{2} \mbox{ for }-R\leq t\leq R$

6)Thus $f\left( e^{\pi i/4}t+\frac{1}{2} \right) =$ $\exp \left\{ i \pi \left( e^{\pi i/4 }t+ \frac{1}{2} \right) ^2 \right\} =$ $e^{ -\pi t^2+\pi ict+i\pi/4}$

7)Similarly the left sloped parametrization gives $f\left( e^{\pi i/4} - \frac{1}{2} \right) = e^{ - \pi t^2 - \pi i c t + i\pi /4}$

8)Thus, $g\left( e^{i\pi /4} t + 1/2 \right) = \frac{ e^{ -\pi t^2+\pi i e^{i\pi /4}t+i\pi/4} }{\cos \pi e^{i\pi /4 t} } \mbox{ and } g\left( e^{i\pi /4} t - 1/2 \right) = \frac{ e^{ - \pi t^2 - \pi i e^{i\pi/4} t + i\pi /4} }{\cos \pi e^{i\pi /4} t}$

9)These two sloped contours add up to
$\int_{-R}^R \frac{ \exp (-\pi t^2) \exp (i\pi /4)}{\cos \pi e^{\pi i/4} t} \left( e^{i\pi e^{\pi i/4}t} + e^{-i\pi e^{\pi i/4} t} \right) e^{i\pi /4} dt$ $= \int_{-R}^R \frac{ 2e^{\pi i/2} \exp(-\pi t^2)\cos \pi e^{\pi i/4} t}{\cos \pi e^{\pi i/4} t} dt$ $= 2i \int_{-R}^R e^{-\pi t^2} dt$

10)There are still the two upper horizontal sections.

11)If we look at the upper one we have for all $s\in [-1/2,1/2]$:
$\left| \exp \left[ i\pi (Re^{i\pi/4}+s)^2 \right] \right| = \exp \left[ -\pi (R^2+\sqrt{2} R s) \right] \leq \exp [-\pi (R^2 - R/\sqrt{2} ) ]$
And,
$|\sin [\pi (Re^{\pi i/4} + s) ] | = \left| \sin [ (\pi / \sqrt{2}) (R+iR+\sqrt{2} s) \right| \geq \sinh (\pi R/\sqrt{2} )$

12)The above inequalitities imply that the contour integral along the upper horizontal section is,
$\left| \int_{-1/2}^{1/2} g(Re^{i\pi /4} + s) ds \right| \leq \frac{\exp [-\pi (R^2 - R/\sqrt{2} ) ]}{\sinh (\pi R/\sqrt{2} )} \to 0 \mbox{ as }R\to \infty$

13)It can similarly be established that $\left| \int_{-1/2}^{1/2}g(-Re^{-i\pi/4} +\xi ) d\xi \right| \to 0$.

14)By the residue theorem the value of this interal is $2\pi i(1/\pi) = 2i$

15)Making $R\to \infty$ we get:
$(PV)\int_{-\infty}^{\infty} e^{-\pi t^2} dt = 1$ if we equate real and imaginary parts.

16)It remains to show that the above integral is convergent in order to removed the Cauchy Principal Value and hence:
$\int_{-\infty}^{\infty} e^{-\pi t^2} dt=1$
Make the substitution $x=\sqrt{\pi} t$ to get,
$\int_{\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

NOTE: In #13 I used $\xi$ because that is what those Show Off Mathematical Physicists like to use.

This is Mine 64th Post!!!

6. Now, define an xy plane. We are integrating over the whole of the xy plane. I am going to choose to do this with polar, rather than rectangular, coordinates. So define:
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$
in the usual manner.

Then $dx dy = r dr d\theta$ and $x^2 + y^2 = r^2$. So...
$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} r dr d\theta$

shouldnt the intergral be

2pi infinity
(intergral) (intergral) e^(-r)^{2} rdrdO
0 -infininty

instead of the other way around for the intergrals as the dO is between 2pi and 0?

and i know the next intergration is trival? but how would it be done?

7. Originally Posted by turion645344
Now, define an xy plane. We are integrating over the whole of the xy plane. I am going to choose to do this with polar, rather than rectangular, coordinates. So define:
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$
in the usual manner.

Then $dx dy = r dr d\theta$ and $x^2 + y^2 = r^2$. So...
$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx dy = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} r dr d\theta$

shouldnt the intergral be

2pi infinity
(intergral) (intergral) e^(-r)^{2} rdrdO
0 -infininty

instead of the other way around for the intergrals as the dO is between 2pi and 0?

and i know the next intergration is trival? but how would it be done?
Hmmm... You are probably right about the integration limits. I'll have to double check my Calc book on that but it does make sense to have the inside integral limits match with the first listed d(variable). I suppose I might have been getting a bit sloppy. If so, my apologies.

I'll write it out again like this so there's no possible confusion:
$I^2 = \int_0^{\infty} \left ( \int_0^{2\pi} e^{-r^2} r d \theta \right ) dr$

Now, for the $\theta$ integral. Note that the integrand does not depend on $\theta$, so it is a constant with respect to the integration. I'm going to pull the "constant" outside of the $\theta$ integral:
$I^2 = \int_0^{\infty} \left ( \int_0^{2\pi} e^{-r^2} r d \theta \right ) dr = \int_0^{\infty} re^{-r^2} \left ( \int_0^{2\pi} d \theta \right ) dr$

The $\theta$ integral is just
$\int_0^{2\pi} d \theta = \theta |_0^{2 \pi} = (2 \pi) - (0) = 2 \pi$, so the integral becomes:
$I^2 = \int_0^{\infty} re^{-r^2} \left ( \int_0^{2\pi} d \theta \right ) dr =
\int_0^{\infty} re^{-r^2} (2 \pi) dr = 2 \pi \cdot
\int_0^{\infty} re^{-r^2} dr$

-Dan

8. Here's another solution:

Define $I=\int_0^\infty e^{-x^2}\,dx.$ Now consider the following integral $\int_0^\infty e^{-ux^2}\,dx,$ for a suitable $u$ constant.

Define $y=\sqrt ux$ for the last integral, so

$
\int_0^\infty {e^{ - ux^2 } \,dx} = \frac{I}
{{\sqrt u }} \implies e^{ - u} \left( {\int_0^\infty {e^{ - ux^2 } \,dx} } \right) = I \cdot \frac{{e^{ - u} }}
{{\sqrt u }}.
$

Integrating both sides with respecto to $u,$ yields

$
\int_0^\infty {\left( {\int_0^\infty {e^{ - ux^2 } \,dx} } \right)e^{ - u} \,du} = I\int_0^\infty {\frac{{e^{ - u} }}
{{\sqrt u }}\,du}=2I^2.
$

The double integral of the LHS can be computed as follows:

$
\int_0^\infty {\int_0^\infty {e^{ - u\left( {x^2 + 1} \right)} \,dx} \,du},
$
now reverse the integration order $
\int_0^\infty {\int_0^\infty {e^{ - u\left( {x^2 + 1} \right)} \,du} \,dx},
$
so

$
\int_0^\infty {\frac{1}
{{1 + x^2 }}\,dx}=\frac{\pi }
{2}
$

and the conclusion follows $\blacksquare$