Gaussian Integral

**synclastica_86** · September 16th 2009, 11:05 AM

I need to integrate:
$1=\int_{-\infty}^{\infty}Ae^{-\lambda(x-a)^2}dx$
to find A. I tried 2 different methods but yielded different answers.

Using $\int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi}$ :
let $u=\lambda(x-a)$ and $du=\lambda dx$
then
$1=A\lambda^{-1}\int_{-\infty}^{\infty}e^{-u^2}du=\frac{A}{\lambda}\sqrt{\pi}$
thus $A=\frac{\lambda}{\sqrt{\pi}}$

Attempt 2, using $\int_{-\infty}^{\infty}Ae^{\frac{-(x-a)^2}{2c^2}}dx=Ac\sqrt{2\pi}$
Here $A=\frac{1}{c\sqrt{2\pi}}$ ; since $\frac{1}{2c^2}=\lambda$ , we have $c=\frac{1}{\sqrt{2\lambda}}$
Therefore $A=\frac{1}{\frac{1}{\sqrt{2\lambda}}\sqrt{2\pi}}=\ sqrt{\frac{\lambda}{\pi}}$

I thought both formula should be valid as listed from wiki. What am I doing wrong?

**redsoxfan325** · September 16th 2009, 11:22 AM

Originally Posted by synclastica_86

I need to integrate:
$1=\int_{-\infty}^{\infty}Ae^{-\lambda(x-a)^2}dx$
to find A. I tried 2 different methods but yielded different answers.

Using $\int_{-\infty}^{\infty}e^{-u^2}du=\sqrt{\pi}$ :
let $u=\lambda(x-a)$ and $du=\lambda dx$ Here's your problem.
then
$1=A\lambda^{-1}\int_{-\infty}^{\infty}e^{-u^2}du=\frac{A}{\lambda}\sqrt{\pi}$
thus $A=\frac{\lambda}{\sqrt{\pi}}$

Attempt 2, using $\int_{-\infty}^{\infty}Ae^{\frac{-(x-a)^2}{2c^2}}dx=Ac\sqrt{2\pi}$
Here $A=\frac{1}{c\sqrt{2\pi}}$ ; since $\frac{1}{2c^2}=\lambda$ , we have $c=\frac{1}{\sqrt{2\lambda}}$
Therefore $A=\frac{1}{\frac{1}{\sqrt{2\lambda}}\sqrt{2\pi}}=\ sqrt{\frac{\lambda}{\pi}}$

I thought both formula should be valid as listed from wiki. What am I doing wrong?

You said: "let $u=\lambda(x-a)$ and $du=\lambda dx$ "

In this case though, $-u^2 = -\underbrace{\lambda^2}_{bad}(x-a)^2$ . You want $u=\sqrt{\lambda}(x-a)$ and $du=\sqrt{\lambda}\,dx$ .

Fixing this will reconcile the two answers you got.

Thread: Gaussian Integral

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