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I'm trying to prove that $$\int_0^{\frac{\pi}{2}} J_0(x\cos\theta)\cos\theta d\theta =\frac{\sin x}{x}$$ where $x>0$ and $J_0$ is the Bessel function. With power series expansion, evaluating the integral and the binomial expansion I got

$$ \int_0^{\frac{\pi}{2}} J_0(x\cos\theta)\cos\theta d\theta = \sum_{k=0}^\infty \frac{(-1)^k}{(k!)^2} \frac{x^{2k}}{2^{2k}} \int_0^{\frac{\pi}{2}}\cos^{2k}\theta\cos\theta d\theta = \\ = \sum_{k=0}^\infty \frac{(-1)^k}{(k!)^2} \frac{x^{2k}}{2^{2k}} \sum_{j=0}^k \binom{k}{j}\frac{(-1)^j}{2j+1}$$

My goal is the power series expansion of $\frac{\sin x}{x}=\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$ so, rearranging some terms, I need to prove that $$\sum_{j=0}^k\binom{k}{j}\frac{(-1)^j}{k!k!(2j+1)2^{2k}} = \frac{1}{(2k+1)!}$$

Wolfram confirms that this is true but I have no idea how to prove that. I tried to do some simplification $$\sum_{j=0}^k \frac{1}{(2k)!!}\frac{(-1)^j}{j!(k-j)!(2j+1)2^{k}}$$ From there I can use the fact that $(2k)!!(2k+1)!!=(2k+1)!$ to conclude but I still dont know if is it possible to prove that $$\sum_{j=0}^k \frac{(-1)^j}{j!(k-j)!(2j+1)2^{k}}=\frac{1}{(2k+1)!!}$$

Can you please help me?

injo
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    Note that the term-by-term integrals being computed here are Wallis integrals. In particular, the problem here amounts to verifying $$W_{2k+1}=\int_0^{\pi/2} \cos^{2k+1} \theta~d\theta=\frac{2^{2k}(k!)^2}{(2k+1)!}$$ which is proven on the Wiki page using the recurrence relation for Wallis integrals. – Semiclassical Aug 18 '24 at 23:20

1 Answers1

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I tried to work on the $\sum_{j=0}^{k}\binom{j}{k}\frac{(-1)^{j}}{2j+1}$.

Using binomial expansion, we have that $(1+x)^k = \sum_{j=0}^{k}\binom{j}{k}x^j$. To get the $\frac{1}{2j+1}$ factor, we have to do some integration so with this in mind, we have that

$$(1-x^2)^k = \sum_{j=0}^{k}\binom{j}{k}(-1)^jx^{2j}$$

To get the $1$ in place of the $x^{2j}$, we will integrate between $0$ and $1$:

$$\begin{align*}\int_0^{1}(1-x^2)^k dx &= \int_{0}^{1}\sum_{j=0}^{k}\binom{j}{k}(-1)^jx^{2j}dx\\&=\sum_{j=0}^{k}\binom{j}{k}(-1)^j\left[\frac{x^{2j+1}}{2j+1}\right]_{0}^{1}\\&=\sum_{j=0}^{k}\binom{j}{k}\frac{(-1)^{j}}{2j+1}\end{align*}$$

Now to get the value of the integral, you can take a look here (make sure to divide their result by 2)

Finally,

$$\begin{align*}\sum_{j=0}^k\binom{k}{j}\frac{(-1)^j}{k!k!(2j+1)2^{2k}} &= \frac{1}{2^{2k}k!k!}\frac{(2k)!!}{(2k+1)!!}\\ &= \frac{(2k)!!}{(2k)!!(2k)!!(2k+1)!!}\\&=\frac{1}{(2k+1)!}\end{align*}$$

Tell me if it is clear for you ;)

Clem105
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  • I got the sum exactly from that integral (with a substitution $u=\sin\theta$ in the first integral). So the trick was to use the recurrence formula instead of explicit evaluation. Everything is clear, thanks. – injo Aug 19 '24 at 05:31