- #1

loloPF

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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?

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- Thread starter loloPF
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- #1

loloPF

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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

loloPF

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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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- #5

Pseudo Statistic

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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

quantumdude

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loloPF said:I just can't solve this:

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

Is this really a high school calculus problem?

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?

- #7

paul-martin

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- #8

quantumdude

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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.

- #9

quantumdude

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Tom Mattson said: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 [itex]exp(-x^2)[/itex]. Shoulda knowed better.

- #10

loloPF

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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.

- #11

quantumdude

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loloPF said: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.

[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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- #12

loloPF

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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.

- #14

loloPF

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

- #15

JoAuSc

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- #16

loloPF

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Anyway, I have the answer!

I just couldn't find it in textbooks 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*

- #18

loloPF

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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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