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I want to show that $\int x^vJ_{v-1}(x)dx = x^vJ_v(x) + C$.

Now I know the recurrence relations of the Bessel equation/function and the one I need to use is $x^vJ_v(x) = x^vJ_{v-1}(x)$

I'm just thinking to set v as a constant and just integrate with respect to x but the $J_v(x)$ is confusing me...I don't see how that will turn out to be $J_{v-1}(x)$

Omar Antolín-Camarena
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Kennan
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1 Answers1

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The relation to use is \begin{align} \frac{d}{dx} \left[ x^{\nu} J_{\nu}(x) \right] = x^{\nu} J_{\nu-1}(x). \end{align} Integrating both sides with respect to $x$ yields \begin{align} \int \frac{d}{dx} \left[ x^{\nu} J_{\nu}(x) \right] \ dx &= \int x^{\nu} J_{\nu-1}(x) \ dx \end{align} or \begin{align} \int x^{\nu} J_{\nu-1}(x) = x^{\nu} J_{\nu}(x) + \mbox{ constant }. \end{align} which provides the desired result.

Leucippus
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  • hmm so is this correct? $\frac{d}{dx} \left[ x^{\nu} J_{\nu}(x) \right] = \int x^vJ_{v-1}(x)dx$ and $x^vJ_{v-1}(x) = \int x^vJ_{v-1}(x) = x^{\nu} J_{\nu}(x) +C$? – Kennan May 17 '14 at 05:20
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    Using $\int \frac{d}{dx} f(x) dx = f(x)$ leads to the desired result. A line has been added to make clear the integration of both sides as stated. – Leucippus May 17 '14 at 05:26