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I am stuck in a question, and don't know where to start. I have to obtain the Laplace transform of $J_0(t)$, I have to let:

$$a_n=\int_{0}^{\pi}(\sin \theta)^{2n}d\theta$$

And now wish to show that: $$a_n= \frac{(2n)!}{2^{2n}(n!)^2}\pi$$

My idea was: I know that:

$$J_0(t)=\sum_{n=0}^{\infty}\frac{(-1)^n t^{2n}}{(n!)^2 2^{2n}}$$

The Laplace transform is represented by:

$$\mathcal{L}(f)=\int_{0}^\infty e^{-st}f(t)dt$$

But can I just plug in the first $a_n$? I don't think so. But where to start now?

Joe Goldiamond
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  • Related here (see the answer of idris) https://math.stackexchange.com/questions/1026506/integration-i-n-int-0-pi-sin2n-theta-d-theta?noredirect=1&lq=1 –  Feb 26 '18 at 16:15
  • Thanks for the advice and there is a second hint by:math.stackexchange.com/questions/476693/using-residue-theorem-to-evaluate-int-0-pi-sin2n-theta-d-theta/476708#476708 ( Hope I quoted correctly) – Joe Goldiamond Feb 26 '18 at 16:24

2 Answers2

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I do not quite follow your train of thoughts, so I will start from scratch.
Given the following definition of $J_0$ $$ J_0(t)=\sum_{n\geq 0}\frac{(-1)^n t^{2n}}{n!^2 2^{2n}}\tag{1} $$ it is trivial that $J_0$ is an entire function. Since $\mathcal{L}(t^{2n})(s)=\frac{(2n)!}{s^{2n+1}}$ we formally have $$ \mathcal{L}(J_0(t))(s) = \sum_{n\geq 0}\frac{(-1)^n}{s^{2n+1}}\cdot\frac{1}{4^n}\binom{2n}{n} \tag{2}$$ and the RHS of (2) is convergent for any $s>1$, since $\frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}$.
By the extended binomial theorem we have $$ \sum_{n\geq 0}\frac{z^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-z}}\tag{3} $$ for any $|z|<1$, hence $\mathcal{L}(J_0(t))(s) =\frac{1}{\sqrt{1+s^2}}$ for any $s>1$. On the other hand $$ J_0(z)=\frac{1}{\pi}\int_{0}^{\pi}\cos\left(z\sin\theta\right)\,d\theta=\frac{1}{\pi}\text{Re}\int_{0}^{\pi}\exp\left(iz\sin\theta\right)\,d\theta\tag{4} $$ holds for any $z>0$, hence by Fubini's theorem $$ \mathcal{L}(J_0(t))(s) = \frac{1}{\pi}\text{Re}\int_{0}^{\pi}\int_{0}^{+\infty}\exp\left(iz\sin\theta-sz\right)\,dz\,d\theta=\frac{1}{\pi}\text{Re}\int_{0}^{\pi}\frac{d\theta}{s-i\sin\theta}\tag{5} $$ and $\mathcal{L}(J_0(t))(s) =\frac{1}{\sqrt{1+s^2}}$ holds for any $s>0$.

Jack D'Aurizio
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  • I think I understand it partly. But how do you get:$$J_0(z)=\frac{1}{\pi}\int_{0}^{\pi}\cos\left(z\sin\theta\right),d\theta$$ Where does it come from? Is it trivial from the former line? – Joe Goldiamond Feb 26 '18 at 20:51
  • @JoeGoldiamond: there are many ways for proving that the last identity and $(1)$ define the same function. A chance is given by proving that the series and the integral representation fulfill the same second-order differential equation, another chance is given by expanding $\cos(\cdot)$ as a power series and computing $\int_{0}^{\pi}\sin^{2n}(\theta),d\theta$ through De Moivre's formula $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2}$ and the orthogonality relation $\int_{0}^{2\pi}e^{mi\theta}e^{-ni\theta},d\theta=2\pi\delta(m,n)$. – Jack D'Aurizio Feb 26 '18 at 21:32
  • Can you recommend me any literature, preferable on the internet, to get to this $J_0$ equation? – Joe Goldiamond Feb 26 '18 at 22:03
  • @JoeGoldiamond: http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html – Jack D'Aurizio Feb 26 '18 at 22:05
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It is possible to solve this using the DE-definition of Bessel function as well as some basic properties of Laplace transforms.

Recall that the definition of the zeroth order Bessel function $J_0(x)$ is that it satisfies

$$xJ_0''+J_0'+xJ_0=0.$$

Take the Laplace transform, and using a property about derivative of Laplace transform: $$\widehat{f'}=p\widehat{f}-f(0)$$ $$\widehat{f''}=p^2\widehat{f}-pf(0)-f'(0)$$ and a property about derivative of a Laplace transform: $$\widehat{tg}=-\widehat{g}\ '$$

We get the following equation

$$(p^2+1)\widehat{J_0} \ ' + p\widehat{J_0} = 0$$

and thus $\widehat{J_0} = A/\sqrt{1+p^2}$. We can further find $A=1$ using $J_0(0)=1$ and the property that

$$p\widehat{f} \rightarrow f(0) \ \textrm{as} \ p\rightarrow \infty.$$

Benjamin Wang
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