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If JrJ_r is the integral

0xrexp(x2)dx\int_0^{\infty} x^r \exp \left(-x^2\right) d x

show that (a) J2r+1=(r!)/2J_{2 r+1}=(r !) / 2, (b) J2r=2r(2r1)(2r3)(5)(3)(1)J0J_{2 r}=2^{-r}(2 r-1)(2 r-3) \cdots(5)(3)(1) J_0.

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Answered 2 years ago
Answered 2 years ago
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a. We first derive a recurrence relationship for J2r+1J_{2r+1}

J2r+1=0xx2rex2dx=0x2r2(2x)(ex2)dx\begin{aligned} J_{2r+1}&=\int_0^{\infty} xx^{2r}e^{-x^2}dx\\ &=\int_0^{\infty}-\frac{x^{2r}}{2}(-2x)(e^{-x^2})dx\\ \end{aligned}

We can solve the equation using integration by parts.

u=x2r2du=2rx2r12dv=2x(ex2)dxv=ex2\begin{aligned} u&=-\frac{x^{2r}}{2}&&\rightarrow du=-\frac{2rx^{2r-1}}{2}\\ dv&=-2x(e^{-x^2})dx&&\rightarrow v= e^{-x^2} \end{aligned}

Evaluate the integration using :

uvvduuv-\int vdu

J2r+1=x2r2(ex2)0+02rx2r12ex2dx=0+rJ2r1\begin{aligned} J_{2r+1}&=-\frac{x^2r}{2}(e^{-x^2})\Bigg|_0^{\infty}+\int_0^{\infty}\frac{2rx^{2r-1}}{2}e^{-x^2} dx\\ &=0+rJ_{2r-1}\\ \end{aligned}

This gives;

J2r+1=r(r1)1J1J_{2r+1}=r(r-1)\cdots1J_1

For J1J_1:

J1=0xex2dx\begin{aligned} J_1&=\int_0^{\infty} xe^{-x^2}dx\\ \end{aligned}

We can solve using Integration by substitution.

u=x2du=2xdxu=-x^2\rightarrow du=-2xdx

New Boundaries of Integration Upper Boundaries

x=u=x=\infty\longrightarrow u=-\infty

Lower Boundaries

x=0u=0x=0\longrightarrow u=0

Substitute everything to the equation:

J1=0eu2du\begin{aligned} J_1&=\int_0^{-\infty}\frac{-e^u}{2}du\\ \end{aligned}

Switch the order of integration by multiplying the integrand by 1-1.

J1=0eu2du=eu20=120=12\begin{aligned} J_1&=\int_{-\infty}^0\frac{e^u}{2}du\\ &=\frac{e^u}{2}\Bigg|_{-\infty}^0\\ &=\frac{1}{2}-0\\ &=\frac{1}{2} \end{aligned}

Hence, J2r+1=12r!J_{2r+1}=\frac{1}{2}r!

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