show that
$$(a^2-b^2)\int_{0}^{P} J_{v}(ax)J_{v}(bx)x\,dx=P\left\{bJ_{v}(aP)J^{'}_{v}(bP)-aJ^{'}_{v}(ap)J_{v}(bP)\right\}$$
when $$J^{'}_{v}(aP)=\d{J_{v}(ax)}{(ax)},(x=P)$$
(My thinking)
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identities
$$\d{}{x}\left\{ x^{-v}J_{v}(ax)\right\}=-ax^{-v}J_{v+1}(ax)$$
$$\d{}{x}\left\{ x^{v+1}J_{v+1}(ax)\right\}=ax^{v+1}J_{v}(ax) $$
$$\d{}{x}\left\{ x^{v}J_{v}(x)\right\}=x^{v}J_{v-1}(x) $$
$$\d{}{x}\left\{ x^{v+1}J_{v+1}(x)\right\}=x^{v+1}J_{v}(x) $$
$$x^{v+1}J_{v+1}(x)=\int \left\{ x^{v+1}J_{v}(x) \right\} dx $$
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soluion
$$\int J_{v}(ax)J_{v}(bx)x dx = \int \left[ x^{v+1}J_{v}(ax) \right] \left[ x^{-v}J_{v}(bx) \right]dx $$
$$uv-\int v du = \left[ x^{-v}J_{v}(bx) \right] \left[ \frac{x^{v+1}}{a}J_{v+1}(ax) \right]+\frac{b}{a}\int \left[ x^{-v}J_{v+1}(bx) \right] \left[ x^{v+1}J_{v+1}(ax) \right] dx $$
see (by parts again)
$$\int \left[ x^{-v}J_{v+1}(bx) \right] \left[ x^{v+1}J_{v+1}(ax) \right] dx=\int \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v+1}(ax) \right] dx $$
$$\int \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v+1}(ax) \right] dx=uv-\int vdu $$
$$\int \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v+1}(ax) \right] dx=-\frac{1}{a} \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] + \frac{b}{a}\int \left[ x^{v+1}J_{v}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] dx $$
So...
$$\int J_{v}(ax)J_{v}(bx)x dx = \left[ x^{-v}J_{v}(bx) \right] \left[ \frac{x^{v+1}}{a}J_{v+1}(ax) \right]+\frac{b}{a}\left[ -\frac{1}{a} \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] + \frac{b}{a}\int \left[ x^{v+1}J_{v}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] \right] dx $$
$$\int J_{v}(ax)J_{v}(bx)x dx = \frac{a}{a^2}\left[ x^{-v}J_{v}(bx) \right] \left[ x^{v+1}J_{v+1}(ax) \right] -
\frac{b}{a} \frac{1}{a} \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] + \frac{b}{a} \frac{b}{a}\int \left[ x^{v+1}J_{v}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] dx $$
$$a^{2}\int J_{v}(ax)J_{v}(bx)x dx = a\left[ x^{-v}J_{v}(bx) \right] \left[ x^{v+1}J_{v+1}(ax) \right] -
b \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] + b^2 \int J_{v}(bx) J_{v}(ax) xdx $$
$$(a^{2}-b^{2})\int J_{v}(ax)J_{v}(bx)x dx = a\left[ x^{-v}J_{v}(bx) \right] \left[ x^{v+1}J_{v+1}(ax) \right] -
b \left[ x^{v+1}J_{v+1}(bx) \right] \left[ x^{-v}J_{v}(ax) \right] $$
$$(a^{2}-b^{2})\int J_{v}(ax)J_{v}(bx)x dx = ax J_{v}(bx) J_{v+1}(ax) -
bx J_{v+1}(bx) J_{v}(ax) $$
And finally...
$$(a^{2}-b^{2}) \int_{0}^{P} J_{v}(ax)J_{v}(bx)x \,dx = P(a J_{v}(bP) J_{v+1}(aP) - b J_{v+1}(bP) J_{v}(aP)) $$
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