How Do You Solve the Integral of x^2 exp(-2amx^2/h)?

[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.

 

[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'?

 

[Ray Vickson]

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.

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

 

[leroyjenkens]

Thanks guys.

Does this hint at a different choice of u and v'?

Yes, that was clever.