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maths help- calculus

The question is In = int(x^n e^-x^2) and to show that
In=(n-1)/2 In-2

I tried to do parts but I cant integrate e^-x^2

Original post by bruh2132

The question is In = int(x^n e^-x^2) and to show that
In=(n-1)/2 In-2

I tried to do parts but I cant integrate e^-x^2

Splitting the integrand as

x^n

and

e^{-x^2}

won't work.

You need a different split. Think about the derivative of

e^{-x^2}

and see if that suggests a different split.

OP

Original post by ghostwalker

Splitting the integrand as

x^n

and

e^{-x^2}

won't work.

You need a different split. Think about the derivative of

e^{-x^2}

and see if that suggests a different split.

I can't seem to do it I tried splitting it like x^n-1 and xe^-x^2 but cant get anywhere

Original post by bruh2132

I can't seem to do it I tried splitting it like x^n-1 and xe^-x^2 but cant get anywhere

That's the correct split.

So

\displaystyle I_n = \int \left(x^{n-1}\right)\left(xe^{-x^2}\right)\;dx

\displaystyle =\int \left(x^{n-1}\right)\frac{d}{dx}\left(-\frac{1}{2}e^{-x^2}\right)\;dx

Can you take it from there? If not, what's the issue?

OP

Original post by ghostwalker

That's the correct split.

So

\displaystyle I_n = \int \left(x^{n-1}\right)\left(xe^{-x^2}\right)\;dx

\displaystyle =\int \left(x^{n-1}\right)\frac{d}{dx}\left(-\frac{1}{2}e^{-x^2}\right)\;dx

Can you take it from there? If not, what's the issue?

Thanks, I also forgot to add that the limits are infinity to 0, does that mean that the first term of the integral disappears; the limits to infinity of it is 0 and when 0 is plugged in its 0 too?
Thanks a lot btw

Original post by bruh2132

Thanks, I also forgot to add that the limits are infinity to 0, does that mean that the first term of the integral disappears; the limits to infinity of it is 0 and when 0 is plugged in its 0 too?
Thanks a lot btw

If you mean the

\displaystyle [uv]_0^\infty

part, yes, that will come out to zero.

The zero at the bottom is obvious.

At infinity. As you go to infinity,

e^{x^2}

goes to infinity faster than any polynomial, so

x^{n-1}e^{-x^2}

will go to zero as x tends to infinity.

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