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How does $(xJ_0'(x))'+xJ_0(x)=0\implies\int_0^1 x J_0(ax)J_0(bx) dx={bJ_0(a)J_0'(b)-aJ_0(b)J_0'(a)\over{a^2-b^2}}?$ Thanks.

Perhaps int by parts? But how do I get the RHS form?

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1 Answer 1

up vote 5 down vote accepted

$(xJ_0'(x))'+xJ_0(x)=0$ also means

$$(xJ_0'(ax))'+axJ_0(ax)=0$$

which can be found by substituting $x\to ax$.


EDIT: To make the first step more explicit

$$\frac{d}{dx}\left(x \frac{dJ_0(x)}{dx}\right)+xJ_0(x) = 0 \to \frac{d}{d(ax)}\left(ax \frac{dJ_0(ax)}{d(ax)}\right)+axJ_0(ax) = 0$$


Multiplying this by $J_0(bx)$ and integrating gets you:

$$\int_0^1(xJ_0'(ax))'J_0(bx)dx+ a\int_0^1xJ_0(ax)J_0(bx)dx = 0 \; ,$$

and using partial integration on the first term

$$J_0'(a)J_0(b)-b\int_0^1 xJ_0'(ax)J_0'(bx)dx+ a\int_0^1 xJ_0(ax)J_0(bx)dx = 0 \; .$$

Exchanging $a$ and $b$ in this equation, you also have

$$J_0'(b)J_0(a)-a\int_0^1 xJ_0'(ax)J_0'(bx)dx+ b\int_0^1 xJ_0(ax)J_0(bx)dx = 0 \; .$$

Multypling the first equation by $a$ and the second by $b$ and then subtracting both of them gets you the desired result.

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Thanks! I seem to be unable to get the 1st line result -- change of variable. I get $d/dx [(1/a)x(d/dx (J_0(ax)))]+ axJ_0(ax)=0$? because $d/d(ax)=d/dx\times d/d(ax)$? –  kristoff Oct 21 '11 at 11:10
    
Note that $dJ_0(ax)/d(ax) = J_0'(ax)$. –  Raskolnikov Oct 21 '11 at 11:14
    
Thanks again! :) –  kristoff Oct 21 '11 at 11:22

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