1. ## Bessel function integration

1) ∫ x^(-2) * J2(x) dx
where J2(x) is Bessel function of order 2

2) ∫ x* (Jυ(λx))^2 dx
this is a definite integral with
upper index b
lower index a

2. Originally Posted by Aggressive Duck

1) ∫ x^(-2) * J2(x) dx
where J2(x) is Bessel function of order 2

2) ∫ x* (Jυ(λx))^2 dx
this is a definite integral with
upper index b
lower index a
Do youi have limits of integration for the first. The second (a la Maple) is

$\frac{1}{2} b^2 \left( J^2_n(b) - J_{n+1}(b) J_{n-1}(b) \right) - \frac{1}{2} a^2 \left( J^2_n(a) - J_{n+1}(a) J_{n-1}(a) \right)$

3. Thank you

No the first has no limits. Note: It should be solved in terms of Jo and J1

Would you please just provide the steps of the second one, or is it a standard form? Because I just can't deliver it like that. We were told to use Bessel equation in the form
t^2 * Jn(t) = n^2 * Jn(t) - tJ'n(t) - t^2Jn''(t)

4. Originally Posted by Aggressive Duck
Thank you

No the first has no limits. Note: It should be solved in terms of Jo and J1

Would you please just provide the steps of the second one, or is it a standard form? Because I just can't deliver it like that. We were told to use Bessel equation in the form
t^2 * Jn(t) = n^2 * Jn(t) - tJ'n(t) - t^2Jn''(t)
OK. Let me show that

$\int x J_n^2\, dx = \frac{1}{2} x^2 \left(J_n^2 - J_{n+1} J_{n-1} \right)$ and then your answer is just evaluating at the upper and low limits (note, I've set $\lambda = 1$ and all constants of integration to zero)

First, integration by parts $u = J_n^2,\;\;\;dv = x\,dx$

$\int x J_n^2\, dx =
\frac{1}{2} x^2 J_n^2 - \int x^2 J_n J_n '\, dx \;\;\;\;(*)
$

Next, consider the differential equation for $J_n$

$x^2 J''_n + x J'_n + x^2 J_n - n^2 J_n = 0$

Multiply this by $J'_n$ and integrating (again, no constant of integration)

$\int x^2 J'_n J''_n\, dx + \int x J'^2_n\,dx + \int x^2 J_n J'_n\,dx - \int n^2 J_n J'_n\, dx = 0$

Integrating the first term by parts gives

$\frac{1}{2} x^2 J'^2_n - \int x J'^2_n\,dx + \int x J'^2_n\,dx + \int x^2 J_n J'_n\,dx - \int n^2 J_n J'_n\, dx = 0$

and we see cancellation. The last term also integrates

$\frac{1}{2} x^2 J'^2_n + \int x^2 J_n J'_n\,dx - \frac{1}{2} n^2 J^2_n = 0$

Thus,

$\int x^2 J_n J'_n\,dx = \frac{1}{2} n^2 J^2_n- \frac{1}{2} x^2 J'^2_n$

or

$\int x^2 J_n J'_n\,dx = - \frac{1}{2}\left(x J'_n - n J_n \right)\left(x J'_n + n J_n \right)$

Now two properties of the Bessel J function Bessel Function

$J'_n = J_{n-1} - \frac{n}{x} J_n\;\;\;J'_n = \frac{n}{x} J_n - J_{n+1}$

Using these gives

$\int x^2 J_n J'_n\,dx = \frac{1}{2}\left(x J_{n+1} \right)\left(x J_{n-1}\right) = \frac{1}{2}x^2 J_{n+1} J_{n-1}$

and substitution into (*) gives the result.