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# If $J_r$ is the integral$\int_0^{\infty} x^r \exp \left(-x^2\right) d x$show that (a) $J_{2 r+1}=(r !) / 2$, (b) $J_{2 r}=2^{-r}(2 r-1)(2 r-3) \cdots(5)(3)(1) J_0$.

Solution

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a. We first derive a recurrence relationship for $J_{2r+1}$

\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.

\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 :

$uv-\int vdu$

\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;

$J_{2r+1}=r(r-1)\cdots1J_1$

For $J_1$:

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

We can solve using Integration by substitution.

$u=-x^2\rightarrow du=-2xdx$

New Boundaries of Integration Upper Boundaries

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

Lower Boundaries

$x=0\longrightarrow u=0$

Substitute everything to the equation:

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

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

\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, $J_{2r+1}=\frac{1}{2}r!$

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