Tricky integral

[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?

[hotvette]

Try:

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

hotvette

[loloPF]

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

[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

[Pseudo Statistic]

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

[Tom Mattson]

I just can't solve this:
I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx


Is this really a high school calculus problem? :eek:


Anyone got a hint?

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?

[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?

[Tom Mattson]

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.

[Tom Mattson]

Is it permissible to interchange the operations of ln and integration here?

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.

[loloPF]

Lots of answers, thanks guys! :smile: but still no result :cry:

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.

[Tom Mattson]

Tom Mattson: No this is not a high school problem, any recommendation on where to post it?


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. :frown:

[loloPF]

hotvette: Now I see what you meant :smile: 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!

[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

[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 (o:)shame on me!) but nothing good so far.

[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.

[loloPF]

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

Anyway, I have the answer! :smile:
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 :cry: 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.

[hotvette]

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

hotvette

[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 :cool: "... 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)

[hotvette]

thanks. Glad you solved it.