Prove: ∫0∞x2ne−x2dx=(2n)!22nn!π2

James Dale

James Dale

Answered

2022-01-05

Prove:
0x2nex2dx=(2n)!22nn!π2

Answer & Explanation

Bernard Lacey

Bernard Lacey

Expert

2022-01-06Added 30 answers

Alternatively, set
I(α)=0eax2dx
differentiate n times with respect to α and evaluate at α=1
To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that I(α)=12πα and differentiating to obtain
dndanI(a)=(1)n0x2neax2dx
some algebraic manipulation and evaluating at a=1 will yield the wanted identity.
godsrvnt0706

godsrvnt0706

Expert

2022-01-07Added 31 answers

Alternatively, integration by parts works immediately.
an=0x2nex2
Consider U=x2n1 so that du=(2n1)x2n2, and dv=xex2 so that V=12ex2.
Then
0x2nex2dx=12ex2x2n100(2n1)x2n212ex2dx
=2n120x2n2ex2dx=(2n1)2n22n0x2n2ex2dx
Hence
an=(2n)(2n1)22nan1
and since a0=π2 we conclude
an=0x2nex2dx=(2n)!22nn!π2
star233

star233

Expert

2022-01-11Added 238 answers

Let's suppose that, one way or another, you know that ex2dx=π. Then
e2txx2dx=et2e(tx)2dx=et2π
On the other hand,
e2txx2dx=(n02ntnxnn!)ex2dx=n02ntnn!xnex2dx
Finally, note that by evenness,
x2nex2dx=20x2nex2dx

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