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?
Best Answer
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 $s>0$.