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Prove that $$\frac{\sin(x)}{x}=\int_0^\frac{\pi}{2}J_0(x\cos(\theta))\cos(\theta)\,d\theta \tag{a}$$ $$\frac{1-\cos(x)}{x}=\int_0^\frac{\pi}{2}J_1(x\cos(\theta))\,d\theta \tag{b}$$ Hint: $$\int_0^\frac{\pi}{2}\cos^{2s+1}(\theta)\,d\theta = \frac{2\cdot 4\cdot6\cdots(2s)}{1\cdot3\cdot5\cdots(2s+1)}$$

I have no idea how to approach this problem. Any suggestions? I did express sine function in exponential form but then I have no clue where to go from there so that I can end up with the integral as an answer indicated above.

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  • $\begingroup$ I gave you the answer and it is the first answer to be posted. I want you to do the details if you want to learn. $\endgroup$ – Mhenni Benghorbal Aug 30 '14 at 21:34
  • $\begingroup$ ... except that I already gave all the details in the answer of mine :D $\endgroup$ – Jack D'Aurizio Aug 30 '14 at 22:31
  • $\begingroup$ @JackD'Aurizio: The main thing in solving problems is idea! The details are left for the OP's. So Let me deal with the OP instead of jumping and working on my ideas and hints. $\endgroup$ – Mhenni Benghorbal Aug 30 '14 at 23:47
  • $\begingroup$ @MhenniBenghorbal: you are free to do what you want, but I think that to take the exclusive right to use power series and to state that I "jumped on your ideas" is a bit too much. Bury the hatchet, please. $\endgroup$ – Jack D'Aurizio Aug 30 '14 at 23:50
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    $\begingroup$ I can't stand: I have already written a solution. I see no reason to avoid the use of power series just because you have suggested to use power series. I am free to use them just as you. $\endgroup$ – Jack D'Aurizio Aug 30 '14 at 23:55
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Since $$J_0(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{4^m\,m!^2}x^{2m}\tag{1}$$ we have: $$\begin{eqnarray*}\int_{0}^{\pi/2}J_0(x\cos\theta)\cos\theta\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{(2m+1)!}=\frac{\sin x}{x}.\end{eqnarray*}$$ In a similar way $$ J_1(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{2\cdot 4^m\, (m+1)\,m!^2}x^{2m+1}\tag{2}$$ gives: $$\begin{eqnarray*}\int_{0}^{\pi/2}J_1(x\cos\theta)\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m+1}}{(2m+2)!}=\frac{1-\cos x}{x}.\end{eqnarray*}$$ $(1)$ and $(2)$ can be derived from the integral representations given in the Wikipedia page for Bessel functions.

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  • $\begingroup$ I already gave him the hint or you are using some thing different. $\endgroup$ – Mhenni Benghorbal Aug 30 '14 at 21:12
  • $\begingroup$ I just wrote down all the steps. $\endgroup$ – Jack D'Aurizio Aug 30 '14 at 21:14
  • $\begingroup$ The whole idea is to teach people here. So that's why we give the main idea which is the key to solve the problem and let the OP to do the details. Me myself I know how do these details but the as I said it is not of the benefit of the OP. Thanks for doing this. $\endgroup$ – Mhenni Benghorbal Aug 30 '14 at 21:17
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    $\begingroup$ I know, but I thought that an answer like "Use power series. And the identity it was given." would be quite useless for teaching purposes. Sometimes details matter - just my two cents, obviously. $\endgroup$ – Jack D'Aurizio Aug 30 '14 at 21:20
  • $\begingroup$ @JackD'Aurizio I am having difficulty deriving (1) and (2). I have gone through the Wikipedia page and my textbook and attempted to derive these on my own but cannot seem to get them. Can you direct me on how you got these please? $\endgroup$ – Aksel'sRose Aug 4 '16 at 3:22

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