# Higher dimension Integrals

1. Jun 6, 2010

### substance90

1. The problem statement, all variables and given/known data
Calculate the following integrals:

(a) $$I(n,\alpha) = \int_{0}^{\infty} e^{-\alpha x^2}x^n dx$$ for n whole integers and $$n \ge 0$$

Calculate all results till n=5.

Tip: First calculate $$I^2(0,\alpha)$$ and $$I(1,\alpha)$$ and then use this to calculate n>1.

(b) $$I(n)=\int_{0}^{\infty} e^{-x}x^n dx$$ for n whole and half integer where $$n\ge -1/2$$

Calculate all results till n=5

Tip: Calculate I(n) using I(0) and I(-1/2).

2. Relevant equations

3. The attempt at a solution
(а) I managed to do the first part using polar coordinates and substitution $$I(0,\alpha) = \sqrt{\frac{\pi}{\alpha}}$$ but I keep getting 0 for $$n \ge 1$$

For example with 1: $$I(1,\alpha) = \int_0^{\infty} e^{-\alpha x^2} x^1 dx = x \sqrt{\frac{\pi}{\alpha}} -\int_0^{\infty} \sqrt{\frac{\pi}{\alpha}} 2x dx = x^2 \sqrt{\frac{\pi}{\alpha}} - x^2 \sqrt{\frac{\pi}{\alpha}} = 0$$

(b) The first part here is ok: $$I(0) = \int_0^{\infty} e^{-x} dx = -e^{-x}$$ but I tried a thousand times to do I(-1/2) and keep going in circle with integration by parts.

Any ideas would be greatly appreciated!

2. Jun 6, 2010

### vela

Staff Emeritus
What happened to the exponential when you integrated by parts?
You can use the result from part (a) to solve (b), so figure out (a) first.

3. Jun 6, 2010

### substance90

Well, we know from the first part that $$\int_{0}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$$ or am I missing something?

4. Jun 6, 2010

### vela

Staff Emeritus
You're doing things out of order. You can't plug the limits in until the end. Consider a simpler example:

$$\int_0^\infty xe^{-x}\,dx$$

Let f=x and dg=e-x dx. Then df=dx and g=-e-x, and you should get

$$\int_0^\infty xe^{-x}\,dx = \left[-xe^{-x}\right]_0^\infty + \int_0^\infty e^{-x}\,dx$$

But what you're doing is saying

$$g = \int_0^\infty dg = \left[-e^x\right]_0^\infty = 1$$

so that

$$\int_0^\infty xe^{-x}\,dx = -x\cdot 1 + \int_0^\infty 1\,dx$$

See the difference?

Anyway, if you attempt integration by parts properly, you're stuck trying to integrate exp(-ax2) without the limits, which you can't do. Instead, for the n=1 case, try the substitution u=-ax2.

Last edited: Jun 6, 2010
5. Jun 7, 2010

### substance90

Well, I did what you said and I got $$- \frac{1}{2}$$ by substitution in $$I(1,\alpha)= \int_0^{\infty} e^{-\alpha x^2} x^1 dx$$ but now Im stuck on all the other expressions $$n \ge 2$$
The Tip says to get solutions for $$I^2(0,\alpha)$$ and $$I(1,\alpha)$$ and use this information to solve the others to n=5 but I dont see how this helps me any further.

Last edited: Jun 7, 2010
6. Jun 7, 2010

### vela

Staff Emeritus
Recheck your work. The integrand is positive over the interval of integration, so the integral should be positive. Also, the α is missing from the answer.

Now use integration by parts to find I(n,α) for n≥2.

7. Jun 7, 2010

### substance90

I corrected it, there was a mistake in the substitution. So, now I have $$\scriptsize I(1,\alpha) = \frac{1}{2\alpha}$$
Im having real trouble with the integration by parts of I(n,a) for n>1, though.

If I choose $$\scriptsize u=x^2$$ and $$\scriptsize \dot{v} = e^{-\alpha x^2}$$ Im stuck because I cannot integrate $$\scriptsize \dot{v}$$ without transformation to polar coordinates.
If I choose the other way around, I get stuck again because I just get x with a growing exponent in the integral.

8. Jun 7, 2010

### vela

Staff Emeritus
Keep one power of x with the exponential so you can integrate it with a substitution.

9. Jun 7, 2010

### substance90

Im not sure I understand what you mean by that. I cannot get to $$\scriptsize x^1$$.

I tried integration by parts that way:

$$\int_0^{\infty} e^{- \alpha x^2} x^2 dx =$$
$$= \biggl [\frac{x^3}{3} e^{- \alpha x^2} \biggr ]_0^{\infty} + \int_0^{\infty} 2 \alpha x e^{- \alpha x^2} \frac{x^3}{3} dx =$$

Now, if I substitute $$u=-\alpha x^2$$ and $$dx=\frac{du}{-2 \alpha x}$$ I cannot manage to get rid of that nasty $$\scriptsize x^3$$.

In the end, whatever I try I just get $$\scriptsize - \frac{x^3}{3} + 0$$ from the integral.

10. Jun 7, 2010

### vela

Staff Emeritus
What I'm saying is you have choices other than u=xn and dv=exp(-ax2)dx. You don't have to put all of the x's from the xn factor into u and just the exponential into dv.

11. Jun 7, 2010

### substance90

That way I got an answer 1/2a for I(2,a), too. Could this possibly be right?

12. Jun 7, 2010

### vela

Staff Emeritus
No, it's not right. You should find I(n+2) is proportional to I(n).

By the way, your answer for I(0,α) is missing a factor of 1/2.

13. Jun 7, 2010

### substance90

Ive got I(0,a) from my "Mathematical methods in physics" textbook, I thought it was supposed to be correct.

Where do I get the 1/2 factor from?

Last edited by a moderator: May 4, 2017
14. Jun 7, 2010

### vela

Staff Emeritus
The lower limit of the integral in the book is $-\infty$. The lower limit on your integral is 0.

15. Jun 7, 2010

### substance90

I see the logic of that but I still cannot mathematically derive this factor of 1/2...

16. Jun 7, 2010

### vela

Staff Emeritus
The easy way to show this is to note that the integrand is an even function, so you'd have

$$\int_0^\infty e^{-\alpha x^2} dx = \frac{1}{2}\int_{-\infty}^\infty e^{-\alpha x^2} dx$$

You could also follow the derivation in the book with slight modifications. Both the x and y integrals would go from 0 to ∞ instead of from -∞ to ∞. The region of integration would therefore be only the first quadrant rather than the entire xy plane, so when you change to polar coordinates, the angle φ would go from 0 to π/2 instead of 0 to 2π. The end result is a factor of 1/2 in the final answer.

17. Jun 7, 2010

### substance90

Thank you for your time, this really helped. At least now I have a small part from the first part of the first problem. Too bad that there is no time to finish it, I have to submit the homework in 20 minutes.
I`ll still look further into that later today, although it will be too late for points from the homework.