# Conjecture: $\int_0^{\infty}dx\frac{e^{i\alpha\sqrt{x^2+1}}}{\sqrt{x^2+1}}J_1(Qx)=\left(e^{i\alpha}-e^{i\sqrt{{\alpha}^2-Q^2}}\right)/Q$

Here $\alpha>0$, $Q>0$, and $J_1$ is a Bessel function. I'm fairly certain the closed form in the title is accurate for a couple of reasons. First, I've evaluated the integral numerically in Mathematica for many $\alpha$ and $Q$ values, and it matches every time. The second reason is a little more elaborate, and involves looking at the solution to the scalar Helmholtz equation in real space and in Fourier space. Comparing these two forms of the solution, one can show (after some work) that the identity in the title must hold.

But I haven't yet been able to prove it. I've tried various substitutions, including $x = \tan(u)$, differentiating under the integral sign, and using the integral form of the Bessel function, to no avail. Needless to say, Mathematica also chokes on this integral.

Thoughts?

Let $K_{\nu}(z)$ be the modified Bessel function of the second kind of order $\nu$.

Similar to the approach used in Chapter 13, section 47 (page 416) of the classical textbook A Treatise on the Theory of Bessel Functions to evaluate $$\int_{0}^{\infty} J_{\mu}(bt) \, \frac{K_{\nu} \left(a\sqrt{t^{2}+z^{2}} \right)}{(t^{2}+z^{2})^{\nu/2}} \, t^{\mu +1} \, dt,$$ we can use the integral representation $$K_{\nu}(z) = \frac{1}{2} \left(\frac{z}{2} \right)^{\nu} \int_{0}^{\infty} \exp \left(-t- \frac{z^{2}}{4t} \right) \frac{dt}{t^{\nu+1}} , \quad \operatorname{Re} \left(z^{2}\right)>0,$$ to first show that $$I(\beta, Q)= \int_{0}^{\infty} \frac{e^{-\beta \sqrt{1+x^{2}}}}{\sqrt{1+x^{2}}} \, J_{1}(Qx) \, dx = \frac{e^{-\beta} -e^{-\sqrt{\beta^{2}+Q^{2}}}}{Q}, \, \quad \left(\beta >0, \, Q>0 \right). \tag{1}$$

We will use the fact that $$K_{1/2}(z) = \sqrt{\frac{\pi}{2z}} \, e^{-z}.$$

We will also use the fact that for $a,p>0$, \begin{align} \int_{0}^{\infty} J_{1}(a t) e^{-p^{2}t^{2}} \, dt &= \int_{0}^{\infty} e^{-p^{2}t^{2}}\sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{at}{2} \right)^{2m+1} \\ &= \sum_{m=0}^{\infty} \frac{(-1)^m}{m!(m+1)!} \left(\frac{a}{2}\right)^{2m+1} \int_{0}^{\infty} t^{2m+1} e^{-p^{2}t^{2}} \\&= \frac{1}{a} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(m+1)!} \left( \frac{a^{2}}{4p^{2}} \right)^{m+1} \\&= \frac{1}{a} \, \left[1-\exp \left(- \frac{a^{2}}{4p^{2}} \right)\right]. \end{align}

Under the assumption that $\beta >0$ and $Q>0$, we get

\begin{align} I(\beta, Q) &= \sqrt{\frac{2 \beta}{\pi}} \int_{0}^{\infty} \frac{K_{1/2} \left( \beta \sqrt{1+x^{2}}\right)}{\sqrt{1+x^{2}}} \, J_{1}(Qx) \, dx \\ &= \frac{1}{2} \sqrt{\frac{2 \beta}{\pi}} \int_{0}^{\infty} \frac{J_{1}(Qx)}{\sqrt{1+x^{2}}} \int_{0}^{\infty} \frac{1}{t^{3/2}} \left(\frac{\beta\sqrt{1+x^{2}}}{2} \right)^{1/2} \exp \left(-t-\frac{\beta^{2}(1+x^{2})}{4t} \right) \, dt \, dx \\ &= \frac{\beta}{2 \sqrt{\pi}} \int_{0}^{\infty} \frac{1}{t^{3/2}} \exp \left(-t-\frac{\beta^{2}}{4t}\right) \int_{0}^{\infty} J_{1}(Qx) \exp \left(\frac{-\beta^{2}x^{2}}{4t} \right) \, dx \, dt \\ &= \frac{\beta}{2 Q \sqrt{\pi}} \int_{0}^{\infty} \frac{1}{t^{3/2}} \exp \left(-t-\frac{\beta^{2}}{4t}\right) \left[1- \exp \left(\frac{Q^{2}t}{\beta^{2}} \right) \right]\, dt \\ &= \frac{1}{Q} \sqrt{\frac{2\beta }{\pi}} \, K_{1/2}(\beta) - \frac{\beta}{2 Q \sqrt{\pi}} \int_{0}^{\infty} \frac{1}{t^{3/2}} \, \exp \left[- \left(1+\frac{Q^{2}}{\beta^{2}} \right)t - \frac{\beta^{2}}{4t} \right] \, dt \\ &= \frac{e^{-\beta}}{Q} - \frac{\sqrt{\beta^{2}+Q^{2}}}{2Q \sqrt{\pi}} \int_{0}^{\infty} \frac{1}{u^{3/2}} \, \exp \left(-u - \frac{\beta^{2}+Q^{2}}{4u} \right) \, du \\ &= \frac{e^{-\beta}}{Q} - \frac{1}{Q} \frac{\sqrt{2} \sqrt{\beta^{2}+Q^{2}}}{ \sqrt{\pi}} \, K_{1/2} \left(\sqrt{\beta^{2}+Q^{2}} \right) \\ &= \frac{e^{-\beta} - e^{-\sqrt{\beta^{2}+Q^{2}}}}{Q}. \end{align}

Now we need to argue $(1)$ holds for $\operatorname{Re}(\beta) \ge 0$.

Fixing $Q$ and using the principal branch of the square root, the right side of the equation is a holomorphic function for $\operatorname{Re}(\beta) >0$.

And Morera's theorem quickly shows that the left side of the equation is also a holomorphic function for $\operatorname{Re}(\beta) >0$.

(I originally attempted to justify differentiation under the integral sign for all $\operatorname{Re}( \beta) >0$, but that approach led nowhere.)

Then to justify $(1)$ holds for $\operatorname{Re}(\beta)=0$, we can apply the dominated convergence theorem since the integrand is dominated for $\operatorname{Re}(\beta) \ge 0$ by the integrable function $\frac{\left|J_{1}(Qx) \right|}{\sqrt{1+x^{2}}}$.

Replacing $\beta$ with $i b$, $b \in \mathbb{R}$, we get $$\int_{0}^{\infty} \frac{e^{-ib\sqrt{1+x^{2}}}}{\sqrt{1+x^{2}}} \, J_{1}(Qx) \, dx = \frac{e^{-i b} - e^{-\sqrt{Q^{2}-b^{2}}}}{Q}\, , \quad b \in \mathbb{R}, \, Q>0.$$

EDIT:

Numerical approximations of the integral suggest that if $b < -Q$, then $$\sqrt{Q^{2}-b^{2}} =-i \sqrt{b^{2}-Q^{2}}.$$

I'm pretty sure that is has to due with the branch cut for $\sqrt{\beta^{2}+Q^{2}}$ (which runs along the imaginary axis and connects $-iQ$ to $iQ$ through the point at infinity).

Below $-Q$ and just to the right of the branch cut, $\arg(\beta^{2}+Q^{2}) = - \pi$.

• that's a good one. Big (+1) Dec 7 '16 at 16:18