What did I do wrong on this Bessel expansion?

[yungman]

Homework Statement

I cannot get the answer given by the book. The question is:

Using Bessel function of order = 2 to represent f(x):
f(x)=0 for 0<x<1/2 and f(x)=1 for 1/2<x<1.


The Answer given by the book is $$-2\sum_{j=1}^{\infty} \frac{J_{1}(\alpha_{2,j})-2J_{1}(\frac{\alpha_{2,j}}{2})}{\alpha_{2,j}J_{1}(\alpha_{2,j})^{2}}J_{2}(\alpha_{2,j}x)$$

I got everything correct except the deminator where I have

$$\alpha_{2,j}J_{3}(\alpha_{j})^{2}$$

I know there is a way to reduce the order if it start with order of 1, I cannot reduce the order of 3 to 1.

Homework Equations

$$\int x^{-p+1}J_{p}(x)dx=-x^{-p+1}J_{p-1}(x)+C$$ for $$p=2\Rightarrow \int x^{-1}J_{2}(x)dx=-x^{-1}J_{1}(x)+C$$

The Attempt at a Solution

We let $$s=\alpha_{j}x \Rightarrow \frac{ds}{\alpha_{j}}=dx, x=\frac{s}{\alpha_{j}},a=1$$

$$A_{j}=\frac{\int_{0}^{a}xf(x)J_{2}(\lambda_{j}x)dx}{\int_{0}^{a}xJ_{2}(\lambda_{j}x)^{2}dx}=\frac{\int_{\frac{1}{2}}^{1}x \frac{1}{x^{2}} J_{2}(\alpha_{j}x)dx}{\frac{a^{2}}{2}J_{3}(\alpha_{j})^{2}dx}=\frac{2\int_{\frac{\alpha_{j}}{2}}^{\alpha_{j}}\alpha_{j}s^{-1}J_{2}(s)\frac{ds}{\alpha_{J}}}{J_{3}(\alpha_{j})^{2}} = \frac{-2}{J_{3}(\alpha_{j})^{2}}[\frac{J_{1}(s)}{s}]_{\frac{\alpha_{j}}{2}}^{\alpha_{j}}$$

$$A_{j}=\frac{-2}{J_{3}(\alpha_{j})^{2}}[\frac{J_{1}(\alpha_{j})}{\alpha_{j}}-\frac{2J_{1}(\frac{\alpha_{j}}{2})}{\alpha_{j}}] = \frac{-2[J_{1}(\alpha_{j})-2J_{1}(\frac{\alpha_{j}}{2})]}{\alpha_{j}J_{3}(\alpha_{j})^{2}}$$

$$f(x)=\sum_{j=1}^{\infty}A_{j}J_{2}(\lambda_{j}x) = -2\sum_{j=1}^{\infty}\frac{[J_{1}(\alpha_{j})-2J_{1}(\frac{\alpha_{j}}{2})]}{\alpha_{j}J_{3}(\alpha_{j})^{2}}J_{2}(\alpha_{j}x)$$


Thanks for your time and Merry Christmas

Alan

 

[yungman]

Anyone please!