- #1

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[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]

I have tried a trigonometric substitution

[tex]x=\gamma \tan(\theta)[/tex]

but I am not happy with the result :grumpy:.

Anyone got a hint?

- Thread starter loloPF
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- #1

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[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]

I have tried a trigonometric substitution

[tex]x=\gamma \tan(\theta)[/tex]

but I am not happy with the result :grumpy:.

Anyone got a hint?

- #2

hotvette

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Try:

[tex]y=\sqrt{x^2+\gamma^2}[/tex]

*hotvette*

[tex]y=\sqrt{x^2+\gamma^2}[/tex]

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- #3

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[tex]dy=\frac{x}{\sqrt{x^2+\gamma^2}}dx[/tex]

and I don't quite see it happening...

Can you be more precise?

- #4

hotvette

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Looks dirty. :DloloPF said:

[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]

I have tried a trigonometric substitution

[tex]x=\gamma \tan(\theta)[/tex]

but I am not happy with the result :grumpy:.

Anyone got a hint?

Numerical integration time! :rofl:

- #6

Tom Mattson

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Is this really a high school calculus problem?loloPF said:I just can't solve this:

[tex]I(\gamma)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+\gamma^2}}dx[/tex]

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?Anyone got a hint?

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- #8

Tom Mattson

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[tex]I(1)=\int_{-\infty}^{\infty} e^{-\sqrt{x^2+1}}dx[/tex]

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

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Tom Mattson

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Apparently, it isn't. I tried it for the indefinite integral, and the antiderivative I obtained is wrong.Tom Mattson said:Is it permissible to interchange the operations of ln and integration here?

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

- #10

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

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.

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Tom Mattson

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If it's a homework assignment, then it is most likely at the College Level. (In the USA, College=University).loloPF said:Tom Mattson: No this is not a high school problem, any recommendation on where to post it?

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.

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

Then I differentiated with respect to [itex]a[/itex]:

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

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

Then I tried to do a u-substitution:

[tex]u=\sqrt{ax^2+\gamma^2}[/tex]

[tex]du=\frac{ax}{\sqrt{ax^2+\gamma^2}}[/tex]

The idea is to integrate the new integral with respect to [itex]x[/itex], then integrate

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

And then again, there's always numerical integration.

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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

- #13

hotvette

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[tex]\int\frac{ye^{-y}}{\sqrt{y^2-\gamma^2}}dy[/tex]

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

[tex]e^{-y}\sqrt{y^2 - \gamma^2} + \int e^{-y}\sqrt{y^2-\gamma^2}dy[/tex]

but got stuck there.

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I'm trying to crack it with Maple now (shame on me!) but nothing good so far.

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Anyway, I have the answer!

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

hotvette

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

*hotvette*

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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

[tex]\frac{1}{2}!=\frac{\sqrt{pi}}{2}[/tex]

The exact relation is:

[tex]\int_0^{\infty}e^{-\sqrt{x^2+z^2}}dx=2zK_1(z)[/tex]

- #19

hotvette

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thanks. Glad you solved it.

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