Alexmahone said:
Prove that $\displaystyle J_1(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)d\theta$ by showing that the right-hand side satisfies Bessel's equation of order 1 and that the derivative has the value $J_1'(0)$ when $x=0$. Explain why this constitutes a proof.
Hi Alexmahone, :)
Let \(\displaystyle f(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\). By the
Leibniz integral rule we get,
\[f'(x)=\frac{1}{\pi}\int_0^\pi\sin\theta\sin(\theta-x\sin\theta)\,d\theta\]
\[f''(x)=-\frac{1}{\pi}\int_0^\pi\sin^{2}\theta\cos(\theta-x\sin\theta)\,d\theta\]
Substituting these in the left hand side of the Bessel's equation of order one we get,
\begin{eqnarray}
x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)&=&-\frac{x^2}{\pi}\int_0^\pi\sin^{2}\theta\cos(\theta-x\sin\theta)\,d\theta+\frac{x}{\pi}\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\\
&~&+\frac{(x^2-1)}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\
&=&\frac{x^2}{\pi}\int_0^\pi\cos^{2}\theta\cos( \theta-x\sin\theta)\,d\theta+\frac{x}{\pi}\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\\
&~&-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\
\end{eqnarray}
Using
integration by parts on \(\displaystyle\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\) we get,
\begin{eqnarray}
x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)&=&\frac{x}{\pi}\int_0^\pi\cos \theta\cos(\theta-x\sin\theta)\,d\theta-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\
&=&\frac{1}{\pi}\int_0^\pi(x\cos\theta-1)\cos(\theta-x\sin\theta)\,d\theta\\
&=&-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d(\theta-x\sin\theta)\\
&=&-\left.\frac{1}{\pi}\sin(\theta-x\sin\theta)\right|_{\theta=0}^{\theta=\pi}\\
&=&0
\end{eqnarray}
\[\therefore x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)=0\]
Also, \(\displaystyle f'(0)=\frac{1}{\pi}\int_0^\pi\sin^{2}\theta\,d \theta=\frac{1}{2}\)
If we consider the
Taylor expansion of the Bessel function of the first kind, differentiation of term by term is possible since it is a power series. Therefore we get,
\[J'_{1}(0)=\frac{1}{2}\]
Hence,
\[J'_{1}(0)=f'(0)=\frac{1}{2}~~~~~~~~~~~(1)\]
The Bessel's equation should have two linearly independent solutions. If the order of the Bessel equation is an integer then the the two solutions are: Bessel function of the first kind and Bessel function of the second kind. (Read
this.)
In our case the order is 1 and hence the function \(f(x)\) should be linearly dependent to either the Bessel function of the first kind\((J_{1})\) or the Bessel function of second kind\((Y_{1})\). But the Bessel function of second kind is not continuous at \(x=0\) whereas \(f(0)\) is finite. Therefore the only possibility is that \(f(x)\) should be linearly dependent to the Bessel function of the first kind. That is,
\[f(x)=\beta J_{1}(x)\mbox{ where }\beta\in\Re\]
Now it can be easily shown that, \(f(0)=J_{1}(0)=0\). Therefore \(\beta\) could not be found out by substituting \(x=0\). However if we differentiate the above equation and substitute zero,
\[f'(0)=\beta J'_{1}(0)\]
By (1),
\[\beta=1\]
\[\therefore J_{1}(x)=f(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\]
Kind Regards,
Sudharaka.
