# Bessel function of order 1

#### Alexmahone

##### Active member
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.

#### Sudharaka

##### Well-known member
MHB Math Helper
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.

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