How to evaluate this integral containing a Bessel function: $\int_1^2 x^{-2} J_2(x)\,dx$?

Question

$$\int_1^2 x^{-2} J_2(x)\,dx$$

I can't solve this integration problem. I want to develop the expression using Derivation of the Bessel function and Recurrence relations between the Bessel functions and get the answer.

Answer 1

You can evaluate in terms of hypergeometric functions. Recall that

$$J_{\nu}(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{-z^2}{4}\right)$$ Therefore $$x^{-2}J_2(x)=\frac{1}{x^2}\frac{(x/2)^2}{\Gamma(2+1)}~{}_0F_1\left(;2+1~;~\frac{-x^2}{4}\right) \\ =\frac{1}{8}~{}_0F_1\left(;3~;~\frac{-x^2}{4}\right)$$

Now, with the substitution $$-x^2/4=z\\ \mathrm dx=-\mathrm iz^{-1/2}~\mathrm dz$$ we get $$\int x^{-2}J_2(x)\mathrm dx=\frac{-1}{8}\int {}_0F_1\left(;3~;~\frac{-x^2}{4}\right)\mathrm dx \\ =\frac{-\mathrm i}{8}\int z^{1/2-1}~{}_0F_1\left(;3~;~z\right)~\mathrm dz$$ Which from here we know has the primitive $$=\frac{-\mathrm i}{8}\frac{z^{1/2}}{1/2}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~z\right)$$

Putting back our substitution $z=-x^2/4$ this is $$=\frac{-\mathrm i}{8}(1/2)^{-1}\sqrt{\frac{-x^2}{4}}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right) \\ =\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)$$ I.E, that is $$\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)\right)=x^{-2}J_2(x)$$

Hence $$\int_1^2 x^{-2}J_2(x)\mathrm dx=\frac{1}{4}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~-1\right)-\frac{1}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-1}{4}\right)\\ \approx 0.102617$$

Confirmed by Mathematica:

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Mathematica says $${}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg|~z\right)\\=\frac{2~I_0(2\sqrt z)\cdot\big(-1+4z+2\pi z~L_1(2\sqrt z)\big)}{3z}-\frac{2~I_1(2\sqrt z)\cdot\big(-1+2z+2\pi z^{3/2}L_0(2\sqrt z)\big)}{3z^{3/2}}$$

Where $$I_\nu(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{z^2}{4}\right)$$ Is a Modified Bessel function and $$L_\nu(z)=\frac{z~(z/2)^\nu}{\sqrt{\pi}~\Gamma(\nu+3/2)}~{}_1F_2\left(\begin{matrix}1 \\ \frac{3}{2},\nu+\frac{3}{2}\end{matrix}~\bigg|~\frac{z^2}{4}\right)$$ Is a modified Struve function.

But IMO this doesn't qualify as "simpler".