Prove: ∫0∞x2ne−x2dx=(2n)!22nn!π2

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Prove:  ∫0∞x2ne−x2dx=(2n)!22nn!π2

Bernard Lacey · Accepted Answer

Alternatively, set  I(α)=∫0∞e−ax2dx  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)n∫0∞x2ne−ax2dx  some algebraic manipulation and evaluating at a=1 will yield the wanted identity.

godsrvnt0706 · Accepted Answer

Alternatively, integration by parts works immediately.  an=∫0∞x2ne−x2  Consider U=x2n−1 so that du=(2n−1)x2n−2, and dv=xe−x2 so that V=−12e−x2.  Then  ∫0∞x2ne−x2dx=12e−x2x2n−1∣0∞−∫0∞(2n−1)x2n−2−12e−x2dx  =2n−12∫0∞x2n−2e−x2dx=(2n−1)2n22n∫0∞x2n−2e−x2dx  Hence  an=(2n)(2n−1)22nan−1  and since a0=π2 we conclude  an=∫0∞x2ne−x2dx=(2n)!22nn!π2

star233 · Accepted Answer

Let&#039;s suppose that, one way or another, you know that ∫−∞∞e−x2dx=π. Then ∫−∞∞e2tx−x2dx=et2∫−∞∞e−(t−x)2dx=et2π On the other hand, ∫−∞∞e2tx−x2dx=∫−∞∞(∑n≥02ntnxnn!)e−x2dx=∑n≥02ntnn!∫−∞∞xne−x2dx Finally, note that by evenness, ∫−∞∞x2ne−x2dx=2∫0∞x2ne−x2dx

Prove: \int_0^{\infty} x^{2n}e^{-x^2}dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}

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New Questions in Calculus 1