I cannot do the following integrals, its an assignment we should do, someone please help
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
I cannot do the following integrals, its an assignment we should do, someone please help
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
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
$\displaystyle \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 $\displaystyle \lambda = 1$ and all constants of integration to zero)
First, integration by parts $\displaystyle u = J_n^2,\;\;\;dv = x\,dx $
$\displaystyle \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 $\displaystyle J_n$
$\displaystyle x^2 J''_n + x J'_n + x^2 J_n - n^2 J_n = 0$
Multiply this by $\displaystyle J'_n$ and integrating (again, no constant of integration)
$\displaystyle \int x^2 J'_n J''_n\, dx + \int x J'^2_n\,dx + \int x^2 J_n J'_n\,dx - \int n^2 J_n J'_n\, dx = 0$
Integrating the first term by parts gives
$\displaystyle \frac{1}{2} x^2 J'^2_n - \int x J'^2_n\,dx + \int x J'^2_n\,dx + \int x^2 J_n J'_n\,dx - \int n^2 J_n J'_n\, dx = 0$
and we see cancellation. The last term also integrates
$\displaystyle \frac{1}{2} x^2 J'^2_n + \int x^2 J_n J'_n\,dx - \frac{1}{2} n^2 J^2_n = 0$
Thus,
$\displaystyle \int x^2 J_n J'_n\,dx = \frac{1}{2} n^2 J^2_n- \frac{1}{2} x^2 J'^2_n$
or
$\displaystyle \int x^2 J_n J'_n\,dx = - \frac{1}{2}\left(x J'_n - n J_n \right)\left(x J'_n + n J_n \right)$
Now two properties of the Bessel J function Bessel Function
$\displaystyle J'_n = J_{n-1} - \frac{n}{x} J_n\;\;\;J'_n = \frac{n}{x} J_n - J_{n+1} $
Using these gives
$\displaystyle \int x^2 J_n J'_n\,dx = \frac{1}{2}\left(x J_{n+1} \right)\left(x J_{n-1}\right) = \frac{1}{2}x^2 J_{n+1} J_{n-1}$
and substitution into (*) gives the result.