- #1

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I have to find a general formula for the function

1/(2n)![tex]\int_{-\inf}^{\inf}x^{2n}*e^{-ax^2} [/tex]

I am a little bit lost in how to proceed - any hints appreciated thanks in advance

- Thread starter greisen
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- #1

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I have to find a general formula for the function

1/(2n)![tex]\int_{-\inf}^{\inf}x^{2n}*e^{-ax^2} [/tex]

I am a little bit lost in how to proceed - any hints appreciated thanks in advance

- #2

tiny-tim

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Hi greisen!1/(2n)![tex]\int_{-\infty}^{\infty}x^{2n}*e^{-ax^2} [/tex]

(btw, it's "\infty")

I assume you've tried integrating by parts, and found that if you start with the x^2n, it just gets worse, and you can't start with the e^{-ax^2}, because it doesn't have a nice anti-derivative?

Hint: can you split the product in some

- #3

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yes I have tried integration by parts but that got very messy.

I don't quite understand how to rewrite the equation involving the anti-derivative?

Best,

- #4

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[tex]\int_{-\infty}^{\infty}x^{2n}*e^{-ax^2} = \int_{-\infty}^{\infty}x^{2n-1}xe^{-ax^2}dx [/tex] then performing integration by parts by letting

[tex]u=x^{2n-1},v=\int xe^{-ax^2}dx[/tex]

I think this would work. By performing integ by parts a couple of times, i think you will be able to notice some pattern, then use induction to prove the general case.

P.S. This is basically what tiny-tim's Hint says...i hope at least...lol...

- #5

tiny-tim

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Yutup, sutupidmath is right, of course.

The trick is to recognise the difference between a problem and a solution.

In this case, the problem is that you look at ∫e^-ax^2

and think

"I

and change that round to

"I

Problem … solution … it's just a state of mind!

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