This question is a sequel to this previous question. As before, some background information is needed first as follows from my textbook: The standard form of Bessel’s differential equation is $$x^2y^{\prime\prime}+xy^{\prime} + (x^2 – p^2)y=0\tag{1}$$ where $(1)$ has a first solution given by $$\fbox{$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}$}\tag{2}$$ and a second solution given by $$\fbox{$J_{-p}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}$}\tag{3}$$ where $J_p(x)$ is called […]

$$ \int J_0(x)\sin x~{\rm d}x $$ Where $J_0$ is Bessel function of first kind of order $0$ This what I tried $$ \int J_0(x)\sin x~{\rm d}x= -J_0(x) \cos x – \int J_0′(x)\cos x~{\rm d}x $$ $$ J_0′(x)=-J_1(x) $$ $$ \int J_0(x)\sin x ~{\rm}x= -J_0(x) \cos x -(J_1(x)\sin x – \int J_1′(x)\sin x~{\rm d}x) $$ $$ […]

Inspired by this question I’m trying to prove that $$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx e^{\frac{x}{2}}$$ So I needed to find the value of $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{\frac{x}{2}^k}{k!}} \\ = \lim_{m \to \infty} […]

This is the first part of a proof that $J_n(x)N_{n+1}(x)-J_{n+1}(x)N_n(x)=-\dfrac{2}{\pi x}$: Write Bessel’s equation $$x^2y^{\prime\prime}+xy^{\prime} + (x^2 – p^2)y=0\tag{1}$$ with $y=J_p$ and again with $y=J_{-p}$; multiply the $J_p$ equation by $J_{−p}$ and the $J_{−p}$ equation by $J_p$ and subtract to get $$\frac{\mathrm{d}}{\mathrm{d}x}\left[x\left(J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}\right)\right]=0\tag{2}$$ So I wrote $(1)$ with $y=J_p$: $$x^2J_p^{\prime\prime}+xJ_p^{\prime} + (x^2 – p^2)J_p=0\tag{3}$$ and again […]

I need advice on how to solve the following integral: $$ \int_0^\infty J_0(bx) \sin(ax) dx $$ I’ve seen it referenced, e.g. here on MathSE, so I know the solution is $(a^2-b^2)^{-1/2}$ for $a>b$ and $0$ for $b>a$, but I don’t know how to get there. I have tried to solve it by using the integral […]

I’d like to know what the explicit solution to the following integral is: $$\displaystyle \int_{t}^{\infty} \frac{J_{d/2}^{2}(x)}{x} \ \mathrm{d}x,$$ where $t > 0$, $d \in \mathbb{N}$, and $J_{\nu}$ denotes the Bessel function of the first kind. Using Mathematica, I’ve been able to get some results when $d$ is even. For instance, if we take $d = […]

I have come across an infinite series involving Bessel functions where the summation is over the argument inside the Bessel function (rather than over the index of the Bessel function, which seems to be the case usually studied). Specifically I am wondering whether a closed form is known for $$\sum_{n=1}^{\infty} \frac{J_k(nz)}{n^k}.$$ Here, $k$ is an […]

I encountered the following integral in a physical problem $$I=\int r^2 \text{J}_0(\alpha r) \text{I}_1(\beta r)\text{d}r$$ where $\text{J}_0$ is the Bessel function of first kind of order $0$ and $\text{I}_1$ is the modified Bessel function of order $1$. Also, $\alpha$ and $\beta$ are arbitrary real numbers. It seems that MAPLE and WOLFRAM are not able to […]

Taking a small extract of this previous bounty question of mine: It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{1}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are constants. To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{3}$$ If […]

I have two integers $m$ and $n$ and a real $t$. I’m trying to prove the following identity : $$\sum\limits_{j=0}^{+\infty}J_{j+n}(t)J_{j+m}(t)=\frac{t}{2(m-n)}(J_{m-1}(t)J_n(t)-J_m(t)J_{n-1}(t))$$ where $J_n$ is the Bessel function of the first kind. I don’t really know where to start to prove this.

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