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Integral of a product of Bessel functions of the first kind

Mathematics Asked by user740332 on November 29, 2021

I want to do this integral $H(rho)=int_{0}^{infty} J_1(2 pi Lr)J_0(2pi rho r)dr$, where $J_1$ and $J_0$ are Bessel functions of the first kind and $Lin mathbb{R}$ is a constant, so I tried to do this in the Mathematica, but he failed. When I tried to put some value to $L$ and $rho$, the software calculate numerically, so I ploted $H(rho)$ for a fixed $L$ and the result of the plot is a function like $rect(x/L)$, such that

begin{equation}
{displaystyle operatorname {rect} (t)=left{{begin{array}{rl}0,&{text{if }}|t|>{frac {1}{2}}\{frac {1}{2}},&{text{if }}|t|={frac {1}{2}}\1,&{text{if }}|t|<{frac {1}{2}}.end{array}}right.}
end{equation}

I’m not sure about this result for $H(rho)$, so I searched in the internet and didn’t find none property for solve this integral, I don’t know if, in fact $H(rho)=rect(x/L)$ or something of this type. Someone knows if this result is correct? This integral have an analytic solution?

One Answer

The general relationship you need is Eqn. 10.22.63 in the Digital Library of Mathematical Functions: $$ int_{0}^{infty}J_{mu}left(axright)J_{mu-1}left(bxright)mathrm{d}x=% begin{cases}b^{mu-1}a^{-mu},&0<b<a,\ (2b)^{-1},&b=a(>0),\ 0,&0<a<b,end{cases} $$ assuming $Re(mu) > 0$.

In your case, $a = 2 pi L$, $b = 2 pi rho$, and $mu = 1$. The result you found above is what you'd get if you set $a = 1$ and $b = t$.

While the given equation seems to require positive $a$ and $b$, I believe you can extend the result to all real non-zero $a$ and $b$ by changing the variable of integration $t to - t$ and/or using the property that $J_n(-x) = (-1)^n J_n(x)$.

Answered by Michael Seifert on November 29, 2021

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