# The integral $\int \frac{J_{d/2}^{2}(x)}{x} \ \mathrm{d}x$

I'd like to know what the explicit solution to the following integral is:

$$\displaystyle \int_{t}^{\infty} \frac{J_{d/2}^{2}(x)}{x} \ \mathrm{d}x,$$

where $t > 0$, $d \in \mathbb{N}$, and $J_{\nu}$ denotes the Bessel function of the first kind.

Using Mathematica, I've been able to get some results when $d$ is even. For instance, if we take $d = 2$, then Mathematica returns the expression

$$\displaystyle \frac{1}{2}(J_{0}^{2}(t) + J_{1}^{2}(t)),$$

and when $d = 4$, we obtain a similar expression involving polynomials in $t$ multiplied by some factors of $J_{0}$ and $J_{1}$. However, when $d$ is odd, it returns expressions involving Fresnel integrals. This entices the question: is there an explicit solution to the above integral? If $d$ is even then the integral solutions (according to Mathematica) look like they can be given by some kind of recurrence relation. Does anyone know what it is?

I managed to find the following formula here:

$$\displaystyle \int \frac{J_{d/2}^2(x)}{x} \ \mathrm{d}x = \frac{x^d}{2^dd\ \Gamma^{2}\left(\frac{d+2}{2}\right)} \ _{2}F_{3}\left(\frac{d+1}{2}, \frac{d}{2}; \frac{d+2}{2}, \frac{d+2}{2}, d+1; -x^2 \right)$$

However, Mathematica suggests a closed form is available when $d$ is even, but cannot evaluate those sums. I've also had a look in some integral tables but they don't appear to have this particular integral for general $d$.

I had Mathematica do the following integral, it states $$I_{ad}=\int_t^\infty \frac{J_{a/2}(x)J_{d/2}(x)}{x}\; dx =\frac{J_{\frac{a}{2}}(t) \left(2 (a-d) J_{\frac{d}{2}}(t)+4 t J_{\frac{d+2}{2}}(t)\right)-4 t J_{\frac{a+2}{2}}(t) J_{\frac{d}{2}}(t)}{(a-d) (a+d)}$$ the denominator $(a-d)$ kills it if $a=d$, however if we expand this out we have $$I_{ad} = \underbrace{-\frac{2 d J_{\frac{a}{2}}(t) J_{\frac{d}{2}}(t)}{(a-d) (a+d)}+\frac{2 a J_{\frac{a}{2}}(t) J_{\frac{d}{2}}(t)}{(a-d) (a+d)}}_{A}\underbrace{-\frac{4 t J_{\frac{a+2}{2}}(t) J_{\frac{d}{2}}(t)}{(a-d) (a+d)}+\frac{4 t J_{\frac{a}{2}}(t) J_{\frac{d+2}{2}}(t)}{(a-d) (a+d)}}_{B}$$ we can see that the numerators of $B$ will be the same when $a=d$, as well as the denominator. If we do the limit as $a\to d$ of this expression we get $$I_{dd}= \frac{-t J_{\frac{d}{2}}(t) J^{(1,0)}_{\frac{d}{2}+1}\left(t\right)+t J_{\frac{d}{2}+1}(t) J^{(1,0)}_{\frac{d}{2}}\left(t\right)+J_{\frac{d}{2}}(t){}^2}{d}$$ where $J^{(1,0)}_n(t)$ means the derivative of $J_n(t)$ with respect to $n$. This seems to agree numerically with the original integral of many real values of $t$ and $d$. If we let $d=2$ and simplify we get $$I_{22}=\frac{1}{2}(1-J^2_0(t)-J^2_1(t))$$ which seems to be the correct value rather than the one quoted in the question.

You can get the series representation of $J^{(1,0)}_n(t)$, by differentiating that of $J_n(t)$ $$J^{(1,0)}_m(x)=\sum _{l=0}^{\infty } \frac{(-1)^l 2^{-2 l-m} x^{2 l+m} \left(-H_{l+m}+\log \left(\frac{x}{2}\right)+\gamma \right)}{\Gamma (l+1) \Gamma (l+m+1)}$$