# Changing argument into complex in the integral of Bessel multiplied by cosine

I got a problem solving the equation below: $$\int_0^a J_0\left(b\sqrt{a^2-x^2}\right)\cosh(cx) dx$$ where $J_0$ is the zeroth order of Bessel function of the first kind.

I found the integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6: $$\int_0^a J_0\left(b\sqrt{a^2-x^2}\right)\cos(cx) dx = \frac{\sin\left(a\sqrt{b^2+c^2}\right)}{\sqrt{b^2+c^2}}.$$

My naive intuition says that I can solve the integral above by changing $c\to ic$, so I would get

$$\int_0^a J_0\left(b\sqrt{a^2-x^2}\right)\cosh(cx) dx = \begin{cases} \frac{\sin\left(a\sqrt{b^2-c^2}\right)}{\sqrt{b^2-c^2}}, & \text{if } b > c\\ \frac{\sinh\left(a\sqrt{c^2-b^2}\right)}{\sqrt{c^2-b^2}}, & \text{if } b < c\\ a, & \text{otherwise.} \end{cases}$$

I am not quite sure about this because it contains substitution into complex number. If anyone could confirm that this is correct/incorrect, it would be much appreciated! Confirmation with some proof will be better.

Many thanks!

$$I=\int_{0}^{a}J_0(b\sqrt{a^2-x^2})\cosh(cx)\,dx = a\int_{0}^{1}J_0(ab\sqrt{1-z^2})\cosh(acz)\,dx$$ The trick is now to expand both $J_0$ and $\cosh$ as Taylor series, then to exploit: $$\int_{0}^{1}(1-z^2)^{n}z^{2m}\,dz = \frac{\Gamma(n+1)\,\Gamma\left(m+\frac{1}{2}\right)}{2\cdot\Gamma\left(m+n+\frac{3}{2}\right)}\tag{2}$$ hence, by assuming $b>c$ and applying twice the Legendre duplication formula: $$\begin{eqnarray*} I &=& a\int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n(ab)^{2n}(1-z^2)^n}{n!^2 4^n}\sum_{m\geq 0}\frac{(ac)^{2m}z^{2m}}{(2m)!}\,dz\\&=&\frac{a}{2}\sum_{n,m\geq 0}\frac{(-1)^n a^{2n+2m}b^{2n}c^{2m}}{4^n}\cdot\frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma(n+1)\Gamma(2m+1)\Gamma\left(m+n+\frac{3}{2}\right)}\\&=&\frac{a}{2}\sum_{n,m\geq 0}\frac{(-1)^n a^{2n+2m}b^{2n}c^{2m}}{4^{n+m}}\cdot\frac{\sqrt{\pi}}{\Gamma(n+1) \Gamma(m+1)\Gamma\left(m+n+\frac{3}{2}\right)}\\&=&\frac{a}{2}\sum_{s=0}^{+\infty}\frac{a^{2s}\sqrt{\pi}\,(c^2-b^2)^s}{\Gamma\left(s+\frac{3}{2}\right)4^s\,\Gamma(s+1)}\\[5pt]&=&\color{red}{\frac{\sin\left(a\sqrt{b^2-c^2}\right)}{\sqrt{b^2-c^2}}}\tag{3}\end{eqnarray*}$$ as you claimed. This, in fact, proves also the Gradshteyn and Ryzhik's formula: we just need to replace the $(-1)^n$ factor in the last lines with $(-1)^{n+m}$.