Problem:
Evaluate in terms of bessel functions:
$$\int_{}^{} x^{-2} J_2(x) dx $$
Attempt Rewriting the integrasl as: $\int_{}^{} x^{-1}*x^{-1}J_2(x) dx $ then using Integration By Parts to get: $$\int_{}^{}x^{-2} J_2(x) = -\frac{1}{x^2}J_1 -\int_{}^{} \frac{1}{x^3} J_1 dx $$
Integration by parts again:
$$\int_{}^{}x^{-2} J_2(x) = -x^{-2} J_1 +x^{-3}J_0 -3\int_{}^{} x^{-4} J_0 dx $$
Now, I am not sure how to continue. Is there another simplification? Or another approach that can express the integral without using another integral?
Thanks in advance.
Bessel functions have nice properties, so it is usually not hard to find closed-form solutions of integrals that contain them. Let:
$$I=\int{J_2(x)\over x^2}dx \tag 1$$
We use the property that can be found here:
$$ xJ_2(x)=2J_1(x)-xJ_0(x) \tag 2$$ $$ J_2(x)={2J_1(x)-xJ_0(x) \over x} \tag 3$$
Inserting $(3)$ into $(1)$ we obtain:
$$ I= 2 \int {J_1(x) \over x^3}dx - \int {J_0(x) \over x^2}dx \tag 4$$
Next, we use the property that can be found here (section Indefinite integrals):
$$\int{J_1(x) \over x^m} dx = -{J_1(x) \over mx^{m-1}}+{1\over m} \int{J_0(x)\over x^{m-1}}dx \tag 5$$
In our case $m=3$, so we get:
$$I = -{2J_1(x) \over 3x^2} - {1\over 3} \int {J_0(x) \over x^2}dx \tag 6$$
From the same link and under the same section, we use the property:
$$ \int {J_0(x)\over x^n}dx = {J_1(x)\over (n-1)^2 x^{n-2}}- {J_0(x) \over (n-1) x^{n-1}}-{1\over (n-1)^2}\int{J_0(x)\over {x^{n-2}}}dx \tag 7$$
So for $n=2$ we get:
$$ I = -{2 J_1(x)\over 3x^2}-{J_1(x)\over3}+{J_0(x)\over3x}+{1\over3}\int J_0(x)dx \tag 8 $$
This is a well-known integral. If you look here you will find:
$$\int J_0(x)dx={\pi x \over 2}H_0(x)J_1(x)+{x\over 2}(2-\pi H_1(x))J_0(x)+const \tag 9 $$
where $H_0$ and $H_1$ are the Struve functions. So the final solution to your integral is:
$$ I = \Bigg( {\pi x \over 6}H_0(x)-{2\over3x^2}-{1\over3} \Bigg)J_1(x)- \Bigg( {\pi x \over 6}H_1(x)-{x\over3}-{1\over3x} \Bigg)J_0(x) +const \tag{10}$$
which I verified using wolfram (check alternate forms).