# E^x^2 definite integral

1. Jan 12, 2014

### leroyjenkens

I attached the solution from the solution manual of the integral I'm trying to figure out.

$$\int_{-∞}^{∞}x^{2}exp(\frac{-2amx^{2}}{h})$$

The solution of that integral without the x2 in front is $\sqrt{\frac{{\pi}h}{2am}}$

So with the x2 I assumed I needed to do integration by parts.

So taking u = x2, du = 2xdx
And taking dv to be $exp(\frac{-2amx^{2}}{h})$

v = $\sqrt{\frac{{\pi}h}{2am}}$

But v would only equal that in a definite integral. When I'm doing integration by parts, I have an indefinite integral. So I'm kinda stuck here. When I put it into wolfram alpha, I get an error function for the answer to that indefinite integral. Do I put that answer in as v?

Thanks.

#### Attached Files:

• ###### 1.jpg
File size:
5.7 KB
Views:
116
2. Jan 12, 2014

### CAF123

Hi leroyjenkens,
Reexpress $$\int_{-\infty}^{\infty} x^2 \exp(-2amx^2/h) dx = \int_{-\infty}^{\infty} x \cdot x \exp(-2amx^2/h)dx$$ Does this hint at a different choice of u and v'?

3. Jan 12, 2014

### Ray Vickson

You have the wrong u and v, since your dv is not integrable in terms of elementary functions. Try again.

4. Jan 12, 2014

### leroyjenkens

Thanks guys.
Yes, that was clever.