How Can Bessel Functions Be Integrated Using Recurrence Relations?

[alexmahone]

Find $\displaystyle\int x^2J_0(x)$ in terms of higher Bessel functions and $\displaystyle\int J_0(x)$.

 

[Mathwebster]

$J_0(x)$ satisfies

$x^2 J_0'' + x J_0' + x^2J_0 = 0$

Integrating gives

$\displaystyle \int \left(x^2 J_0'' + xJ_0'\right) dx + \int x^2 J_0dx = c$

or

$\displaystyle x^2 J_0' - x J_0 + \int J_0 dx + \int x^2 J_0 dx = c$

then use the fact the $J_0' = - J_1$

 

[chisigma]

Using the general formula...

$\displaystyle \int x^{n}\ J_{n-1}(x)\ dx = x^{n}\ J_{n}(x)$ (1)

... and taking into account that $\displaystyle J^{'}_{0}(x)= - J_{1}(x)$ , integration by parts gives You ...

$\displaystyle \int x^{2}\ J_{0}(x)\ dx = \int x\ x\ J_{0}(x)\ dx= x^{2}\ J_{1}(x) - \int x\ J_{1}(x)\ dx = x^{2}\ J_{1}(x) + x\ J_{0}(x) - \int J_{0}(x)\ dx$ (2)

Kind regards

$\chi$ $\sigma$

 

[alexmahone]

Thanks.

 

[blue_raver22]

Integrating Bessel functions can be a challenging task, but it often leads to interesting results and connections between different mathematical concepts. In this case, we are asked to find the integral of $x^2J_0(x)$, where $J_0(x)$ is the Bessel function of the first kind with order 0.

To solve this integral, we can use the well-known recurrence relation for Bessel functions: $J_{n+1}(x) = \frac{2n}{x}J_n(x) - J_{n-1}(x)$. Using this relation, we can rewrite the integral as follows:

$\displaystyle\int x^2J_0(x) = \frac{2}{x}\int xJ_1(x) - \int J_0(x)$

We can further simplify the integral by using another recurrence relation: $xJ_n(x) = \frac{1}{2}\left(x^nJ_n(x)\right)'$. Applying this to our integral, we get:

$\displaystyle\int x^2J_0(x) = \frac{2}{x}\left(\frac{1}{2}\left(x^2J_1(x)\right)' - \int J_1(x)\right) - \int J_0(x)$

We can now use the well-known integral of $J_1(x)$ to simplify the expression further:

$\displaystyle\int x^2J_0(x) = \frac{2}{x}\left(\frac{1}{2}\left(x^2J_1(x)\right)' - \frac{J_0(x)}{x}\right) - \int J_0(x)$

Finally, we can use the recurrence relation once again to express $J_1(x)$ in terms of $J_0(x)$ and $J_2(x)$:

$\displaystyle\int x^2J_0(x) = \frac{2}{x}\left(\frac{1}{2}\left(x^2\left(\frac{2}{x}J_1(x) - J_0(x)\right)\right)' - \frac{J_0(x)}{x}\right) - \int J_0(x)$

Simplifying the expression, we get:

$\displaystyle\int x^2J_0(x) = J_2(x)