# Tricky integral

1. Sep 27, 2005

### loloPF

I just can't solve this:
$$I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx$$
I have tried a trigonometric substitution
$$x=\gamma \tan(\theta)$$
but I am not happy with the result :grumpy:.
Anyone got a hint?

2. Sep 27, 2005

### hotvette

Try:

$$y=\sqrt{x^2+\gamma^2}$$

hotvette

Last edited: Sep 27, 2005
3. Sep 27, 2005

### loloPF

Then
$$dy=\frac{x}{\sqrt{x^2+\gamma^2}}dx$$
and I don't quite see it happening...
Can you be more precise?

4. Sep 27, 2005

### hotvette

Just go a bit further. You can get dx in terms of dy, y, and $$\gamma$$ by using the substitution and some algebra. By the way, the suggestion was to simplify the expontential. Hopefully, the resulting integrand will be more reasonable to deal with, such as integration by parts. I haven't solved it, but the substitution looks like it takes you in a promising direction.

hotvette

5. Sep 27, 2005

### Pseudo Statistic

Looks dirty. :D
Numerical integration time! :rofl:

6. Sep 27, 2005

### Tom Mattson

Staff Emeritus
Is this really a high school calculus problem?

How about taking the natural log of both sides, integrating, and then exponentiating the result? Is it permissible to interchange the operations of ln and integration here?

7. Sep 27, 2005

### paul-martin

Confusing the function is govern by gamma, but you are doing an intergration of x. Do you mean that the function is govern both by x and gamma?

8. Sep 27, 2005

### Tom Mattson

Staff Emeritus
No, the x is integrated out. To see this, consider the case $\gamma=1$:

$$I(1)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+1}}dx$$

When that integral is evaluated, you will get a number, not a function of x.

9. Sep 27, 2005

### Tom Mattson

Staff Emeritus
Apparently, it isn't. I tried it for the indefinite integral, and the antiderivative I obtained is wrong.

Actually it should have been obvious to me that this won't work, otherwise someone would have used it to find an antiderivative of $exp(-x^2)$. Shoulda knowed better.

10. Sep 27, 2005

### loloPF

Lots of answers, thanks guys! but still no result

hotvette: I am not sure about this but I'll try. The pain is you have to split the integral because of the term $$\sqrt{y^2-\gamma^2}$$

Pseudo Statistic: I do not want a numerical result this time... sorry.

Tom Mattson: No this is not a high school problem, any recommendation on where to post it? Good try with the log but integrals unfortunately do not have this property, I can not even think of any non trivial integral (that is the integral of 1 or that of 0) that would have this property.

11. Sep 27, 2005

### Tom Mattson

Staff Emeritus
If it's a homework assignment, then it is most likely at the College Level. (In the USA, College=University).

I had another attempt that didn't work, but maybe you can play with it to get something. The trick is called "parameterizing the integral".

I started by generalizing your function to include a second variable.

$$F(\gamma,a)=\int_{-\infty}^{\infty} e^{-\sqrt{ax^2+\gamma^2}}dx$$

Then I differentiated with respect to $a$:

$$\frac{\partial F(\gamma,a)}{\partial a}=\int_{-\infty}^{\infty}\frac{\partial}{\partial a} e^{-\sqrt{ax^2+\gamma^2}}dx$$

$$\frac{\partial F(\gamma,a)}{\partial a}=\int_{-\infty}^{\infty}\frac{-a}{\sqrt{ax^2+\gamma^2}} e^{-\sqrt{ax^2+\gamma^2}}dx$$

Then I tried to do a u-substitution:

$$u=\sqrt{ax^2+\gamma^2}$$
$$du=\frac{ax}{\sqrt{ax^2+\gamma^2}}$$

The idea is to integrate the new integral with respect to $x$, then integrate that with respect to $a$, and then recognize that $I(\gamma)=F(\gamma,1)$.

Of course, I couldn't get it to work because of that $x$ in my $du$. You can solve for $x$ in terms of $u$, but it gets ugly again. But maybe you can find a parameterization that works.

And then again, there's always numerical integration.

Last edited: Sep 27, 2005
12. Sep 27, 2005

### loloPF

hotvette: Now I see what you meant I am taking that direction.

Tom Mattson: Funny you would think of that, I was also looking into finding a differential equation for which "my" integral would be a solution.

I'll let you know guys, I've got to break this one!

13. Sep 27, 2005

### hotvette

Using the substitution $$y = \sqrt{u^2+\gamma^2}$$ I as able to come up with:

$$\int\frac{ye^{-y}}{\sqrt{y^2-\gamma^2}}dy$$

which seems a bit more reasonable. I then tried integration by parts, which resulted in:

$$e^{-y}\sqrt{y^2 - \gamma^2} + \int e^{-y}\sqrt{y^2-\gamma^2}dy$$

but got stuck there.

hotvette

14. Sep 28, 2005

### loloPF

That's where I stand too... noting that given the bounds the first term is zero.
I'm trying to crack it with Maple now (shame on me!) but nothing good so far.

15. Sep 28, 2005

### JoAuSc

So far, I've figured out that since the integrand is an even function, instead of integrating from negative infinity to positive infinity you can integrate from 0 to positive infinity and multiply the result by two. Also, when gamma = 0, it's pretty easy to show that I(0) = 2*1 = 2. I haven't figured out the rest, though.

16. Sep 28, 2005

### loloPF

Thanks JoAuSc, it's a very wise thing to search for simplifications and particular cases. The "rest" is the tough part, though.

I just couldn't find it in text books but this integral is well known as the "modified Bessel function of second kind". As expected it does not have a primitive but that's all right because it is the answer to my original question.

Thanks all for helping out and sorry I had you sweat on such a tough problem.

17. Sep 28, 2005

### hotvette

Interesting. I'm curious as to how you found out. Can you post a web link?

hotvette

18. Sep 29, 2005

### loloPF

A friend a mine is a researcher in Maths and eventhough he is not a specialist in integration, a friend of his just recognised the function right away! Like: "Hey dude, this looks like $$K_1(\gamma)$$, one of the modified Bessel function of second kind "... nevermind.

Anyway you can find some information here:
http://en.wikipedia.org/wiki/Bessel_function
and some more there:
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html

As this function is not usually defined by its integral (but by a more generic differential equation), you might not recognize it at first sight. The one we have actually corresponds to equation 7 in the second link, with n=1 and knowing that
$$\frac{1}{2}!=\frac{\sqrt{pi}}{2}$$

The exact relation is:
$$\int_0^{\infty}e^{-\sqrt{x^2+z^2}}dx=2zK_1(z)$$

19. Sep 29, 2005