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I'm trying to solve problem 3.3 from Jackson's Classical Electrodynamics, but I'm encountering some troubles solving $$ \int_0^R \frac{r^{l+1}}{\sqrt{R^2 - r^2}}\text dr, \qquad l = 0,2,4,6,\ldots $$

As far I know, this integral equals to $$R^{2l+1} \frac{(2l)!!}{(2l+1)!!},$$ where $l!! = l(l-2)!! = l(l-2)(l-4)!!$ is a double factorial.

First problem: this integral doesn't appear in my integrals table (could you recommend one?). Second problem: I tried the substitution $r = R\sin t$ but it didn't help.

So, can you tell a nice substitution to solve this or the name of a book that may have this particular integral? Please don't use the beta function in the answer you may give, I don't know how to use it.

UPDATE Using substitutions $r = Rx$ (as suggested) and $\xi = \sqrt{1 - x^2}$, $$ \int_0^R \frac{r^{l+1}}{\sqrt{R^2 - r^2}}\text{d}r = R^{l+1}\int_0^1 \frac{x^{l+1}}{\sqrt{1 - x^2}}\text{d}x = R^{l+1}\int_0^1 (1 - \xi^2)^{l/2}\text{d}\xi $$

Giving values $l = 0,2,4,6,\ldots$ ($l$ even) I managed to deduce the 2nd equation of the post. So the question can be consider answered

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    $\begingroup$ Try integration by parts to get a recurrence relation. $\endgroup$
    – user7777777
    Commented May 6 at 19:46
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    $\begingroup$ For $\ell > 0$, integration by parts yields $$ \int_0^R r^{\ell} \cdot \frac{r}{R^2 - r^2}\,dr = \ell \int_0^R r^{\ell - 1}\sqrt{R^2 - r^2}. $$ Perhaps that's helpful $\endgroup$
    – Ben Grossmann
    Commented May 6 at 19:49
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    $\begingroup$ Note that $\int_0^R \frac{r^{2l+1}}{\sqrt{R^2 - r^2}}dr =R^{2l+1}I_{2l+1}$, with $$I_{2l+1}= \int_0^1 \frac{r^{2l+1}}{\sqrt{1- r^2}}dr =\frac{2l}{2l+1}I_{2l-1} = \frac{2^l l!}{(2l+1)!!} I_0$$ $\endgroup$
    – Quanto
    Commented May 6 at 20:23
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    $\begingroup$ It's the general version of integration by parts for any $\ell$, not the final answer. $\endgroup$
    – aschepler
    Commented May 7 at 15:45
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    $\begingroup$ We don't really "close" answered questions. You can click the checkmark by someone's answer to select it as most helpful. Or you can write your own answer with the method you ended up using, and then checkmark that. $\endgroup$
    – aschepler
    Commented May 7 at 16:24

3 Answers 3

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Making the substitution $r=Rx^{1/2}$, the integral becomes \begin{align} \int_0^R \frac{r^{l+1}}{\sqrt{R^2 - r^2}}\,dr &=\int_0^1\frac{R^{l+1}x^{(l+1)/2}}{\sqrt{R^2-R^2x}}\,\frac{R\,dx}{2\sqrt{x}} \\ &=\frac{R^{l+1}}{2}\int_0^1 x^{l/2}(1-x)^{-1/2}\,dx \\ &=\frac{R^{l+1}}{2}\,B\left(\frac{l}{2}+1,\frac{1}{2}\right), \tag{1} \end{align} where $$ B(z_1,z_2)=\int_0^1t^{z_1-1}(1-t)^{z_2-1}\,dt=\frac{\Gamma(z_1)\Gamma(z_2)} {\Gamma(z_1+z_2)} \tag{2} $$ is the Beta function.

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Let $r=R\sin\theta$, then $$\int_0^R \frac{r^{l+1}}{\sqrt{R^2-r^2}} d r =R^{l+1} \int_0^{\frac{\pi}{2}} \sin ^{l+1} \theta d \theta $$

For the integral $I_n=\int_0^{\frac{\pi}{2}} \sin ^n x d x$, $$ \begin{aligned} I_n&=-\int_0^{\frac{\pi}{2}} \sin ^{n-1} x d(\cos x) \\ & =(n-1) \int_0^{\frac{\pi}{2}} \cos ^2 x \sin ^{n-2} x d x \\ & =(n-1) \int_0^{\frac{\pi}{2}}\left(1-\sin ^2 x\right) \sin ^{n-2} x d x \\ & =(n-1) I_{n-2}-(n-1) I_n \\ \therefore I_n&=\frac{n-1}{n} I_{n-2} . \end{aligned} $$ by which we can reduce the power by two and so on as

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} \sin ^{2 l+1} \theta d \theta & =I_{2 l+1} \\ & =\frac{2 l}{2 l+1} \cdot \frac{2 l-2}{2 l-1} \cdots \frac{2}{3} \cdot I_1 \\ & =\frac{(2 l)!!}{(2 l+1)!!} \end{aligned} $$ Hence $$ \int_0^R \frac{r^{2 l+1}}{\sqrt{R^2-r^2}} d r=R^{2 l+1} \frac{(2 l)!!}{(2 l+1)!!} $$

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Let $r=x\,R$ then $$I=\int \frac{r^{L+1}}{\sqrt{R^2 - r^2}}\,dr=R^{L+1}\int\frac{ x^{L+1}}{\sqrt{1-x^2}}\,dx$$ Using the Gaussian hypergeometric function $$I=\frac{R^{L+1} }{L+2} \,x^{L+2}\,\, _2F_1\left(\frac{1}{2},\frac{L+2}{2};\frac{L+4}{2};x^2\right)$$and the definite integral is then, after simplifcations, $$I=\int_0^R \frac{r^{L+1}}{\sqrt{R^2 - r^2}}\,dr=\frac{\sqrt{\pi }\,\, R^{L+1}\,\, \Gamma \left(\frac{L+2}{2}\right)}{2 \,\, \Gamma \left(\frac{L+3}{2}\right)}$$

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