# Closed form for $\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx$

Consider the following integral: $$\mathcal{I}(\mu,\nu)=\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx,$$ where $J_\mu(x)$ is the Bessel function of the first kind: $$J_\mu(x)=\sum _{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\mu+1)}\left(\frac x2\right)^{2n+\mu}$$ and $Y_\mu(x)$ is the Bessel function of the second kind: $$Y_\mu(x)=\frac{J_\mu(x)\cos(\mu\pi)-J_{-\mu}(x)}{\sin(\mu\pi)}.$$ I was not able to rigorously establish a closed form for $\mathcal{I}(\mu,\nu)$, but based on numerical integration I made a conjecture: $$\forall\mu,\nu\in\mathbb{R},\hspace{1cm}\mathcal{I}(\mu,\nu)\stackrel{?}{=}\frac{\pi}{2}(\mu^2-\nu^2).$$ Could you please help me to find out if this conjecture is true?

If the conjecture is true, then taking the derivative with respect to $\mu$ at $\mu=1$ we get the following corollary: $$\int_0^\infty\frac{J_0(x)J_1(x)+Y_0(x)Y_1(x)}{J_1(x)^2+Y_1(x)^2}x^{-1}\mathrm dx\stackrel{?}{=}\frac{\pi}{2}.$$
As pointed by O.L., the conjecture is equivalent to $$\int_0^{\infty}\ln\frac{\pi\,x\,H^{(1)}_\nu(x)H^{(2)}_\nu(x)}{2}\mathrm dx\stackrel{?}{=}\frac{\pi\,(4\nu^2-1)}{8},$$ where $H^{(1)}_\nu(x)=J_\nu(x)+i\,Y_\nu(x)$ and $H^{(2)}_\nu(x)=J_\nu(x)-i\,Y_\nu(x)$ are the Hankel functions of the first and second kind.

• Maybe it is worth adding that 1) combinations $J_{\mu}\pm i Y_{\mu}$ are Hankel functions and 2) $J_{1/2}^2(x)+Y_{1/2}^2(x)=\frac{2}{\pi x}$. So the conjecture is equivalent to saying that $$I(\nu)=\int_0^{\infty}\ln\frac{\pi x\,H^{(1)}_{\nu}(x)H^{(2)}_{\nu}(x)}{2}dx=\frac{\pi(4\nu^2-1)}{8}.$$ – Start wearing purple May 21 '13 at 0:44
• I posted a related philosophical question at MO: mathoverflow.net/questions/131528/… – Vladimir Reshetnikov May 22 '13 at 23:09
• Maybe you can use some algebra/functional analysis (something like considering $H_{\nu}^{(1)}(x)$ and $H_{\nu}^{(1)}(x)$ as linearly independent vectors of a certain vector space) to show $I(\nu)$ is a polynomial of degree 2? Then all you'd have to do would be evaluating $I(0)$ and proving $1/2$ and $-1/2$ are roots of $I(\nu)$. – Sylvain Julien May 23 '13 at 17:01
• I added a proof that $H^{(1)}_{\nu}(z)$ doesn't have any zeros in the upper half plane, completing the argument. – The Shadow May 28 '13 at 1:27

I will assume that $\nu$ is real, as in the formulation of the question. Something similar may be true for complex $\nu$ by a similar argument, but there would be an extra complication (perhaps only notational) of working with $\nu$ and its conjugate $\overline{\nu}$ instead of just $\nu$. The Hankel functions $H^{(i)}_{\nu}(z)$ are entire except for a branch cut along the negative real axis. Let $$A^{(1)}_{\nu}(z) = \frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)}, \qquad A^{(2)}_{\nu}(z) = \frac{H^{(2)}_{\nu-1}(z)}{ H^{(2)}_{\nu}(z)},$$ $$B^{(1)}_{\nu}(z) = \frac{H^{(1)}_{\nu+1}(z)}{ H^{(1)}_{\nu}(z)}, \qquad B^{(2)}_{\nu}(z) = \frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)}.$$ We introduce the following notation: for a function $F(z)$ with a branch cut along the negative axis, we let $F(x^{+})$ and $F(x^{-})$ denote the limit of $F(z)$ as $z \rightarrow x$ from the region $\mathrm{Im}(z) > 0$ and $\mathrm{Im}(z) < 0$ respectively. There are identities as follows: $$\text{Eq. 1:} \quad A^{(1)}_{\nu}(x^{+}) = \overline{A^{(2)}_{\nu}(x^{-})} = -A^{(2)}_{\nu}(-x) = - \overline{A^{(1)}_{\nu}(-x)},$$ $$A^{(2)}_{\nu}(x^{+}) = \overline{A^{(1)}_{\nu}(x^{-})}.$$ and the same equations hold for $B$. Note that it's not true that $A^{(2)}_{\nu}(x^{+}) = -A^{(2)}_{\nu}(-x)$, the lack of symmetry here is related to the branch cut. This is an important point. The behavior of $A^{(1)}_{\nu}(z)$ is bad in the region near $x^{-}$, and correspondingly $A^{(2)}_{\nu}(z)$ is bad near $x^{+}$. The Hankel function has nice asymptotic expansions for large $z$. The ratio of such functions at arguments $\nu$ differing by integers is particularly nice, because the complex phase cancels. In particular, one has the following: $$A^{(1)}_{\nu}(z) \sim i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z} + O(z^{-3}),$$ $$A^{(2)}_{\nu}(z) \sim -i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z} + O(z^{-3}),$$ $$B^{(2)}_{\nu}(z) \sim i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z} + O(z^{-3}),$$ $$B^{(1)}_{\nu}(z) \sim -i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z} + O(z^{-3}).$$ This holds outside the bad regions mentioned above. In particular, it holds for $A^{(1)}_{\nu}(z)$ and $B^{(1)}_{\nu}(z)$ for $z$ with argument in $[-\pi + \epsilon,\pi]$, and for $A^{(2)}_{\nu}(z)$ and $B^{(2)}_{\nu}(z)$ with argument in $[-\pi,\pi - \epsilon]$. Let $C_R$ be the semi-circle with centre $0$ and radius $R$ in the upper half plane, oriented anti-clockwise, and thought of as lying above the branch cut in $(-\infty,0])$. Note that this is contained within the range where the asymptotic holds for $A^{(1)}_{\nu}(z)$, and hence $$\lim_{R \rightarrow \infty} \int_{C_R} z \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - i \right) - \frac{(2 \nu - 1)}{2} dz = \pi i \cdot \left( -i \cdot \frac{(4 \nu^2 - 1)}{8} \right) = \frac{\pi (4 \nu^2 - 1)}{8}$$ The main term comes from the residue theorem (applied to a half circle, hence the $\pi i$ rather than $2 \pi i$ factor), and the error term comes from the fact that the integral of $O(z^{-2})$ over a half-circle of radius $R$ and circumference $\pi R$ is $O(R^{-1})$.

We now use the following fact: $H^{(1)}_{\nu}(z)$ has no zeros in the upper half plane. I say that it is a fact, but I couldn't find a reference (Edit: proof of this fact included at the end of this answer). I proved it rigorously by an explicit contour integral computation for various ranges of values of $\nu$, however. (Certainly, by the asymptotic expansion which is valid in the entire upper half plane, it follows that any such zeros, if they exist, must be within some small radius, which one can eliminate by computing $(2 \pi i)^{-1} \oint d \log(f)$.) By the residue theorem (taking $C$ above to be the circle in the upper half plane), we get, for any holomorphic integrand, $$0 = \oint = \int_{C} + \int^{R}_{-R},$$ and hence $$\lim_{R \rightarrow \infty} \int^{R}_{-R} z \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - i \right) - \frac{(2 \nu - 1)}{2} dz = - \frac{\pi (4 \nu^2 - 1)}{8} .$$ Note that the integrand has order $O(1/z)$, so one really has to take the integrand from $-R$ to $R$ and then take the limit for this to make sense. One may apply the same analysis to $H^{(2)}_{\nu}$, except now the zero free region of $H^{(2)}_{\nu}$ is in the lower half plane --- this follows by symmetry from the identity $H^{(1)}_{\nu}(\overline{z}) = \overline{H^{(2)}_{\nu}(z)}$, noting that we are once again in the correct region as far as asymptotics goes. Hence we deduce that $$\lim_{R \rightarrow \infty} \int^{-R}_{R} z \left(\frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} - i \right) - \frac{(2 \nu + 1)}{2} dz = - \frac{\pi (4 \nu^2 - 1)}{8} .$$ Note that the direction of the integral has changed, for orientation reasons.

Warning! There's also another difference between this and the previous integral. The first integral was above the branch cut and this integral is below the branch cut. However, in the first case, we were integrating values of the form $A^{(1)}_{\nu}(x^+)$, which was the good value (in the sense that it was related to three other values by symmetry in equation 1), and here we are integrating $B^{(2)}_{\nu}(x^{-})$, which also is related to three other values by the same equations.

Correcting the order of the second integrand and then subtracting the results, we get $$\lim_{R \rightarrow \infty} \int^{R}_{-R} 1 + z \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - \frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} \right) dz = - 2 \cdot \frac{\pi (4 \nu^2 - 1)}{8}.$$ We now make two observations: the integrand is now $O(z^{-2})$ for large $z$, and hence it exists as a definite integral. Moreover, the integrand is even. In light of the warning, we should really specify that the integrand for values $x \in (-\infty,0]$ is precisely: $$1 + x \left(A^{(1)}_{\nu}(x^{+}) - B^{(2)}_{\nu}(x^{-})\right).$$ (One should check this is the correct function to make the integrand even.) We deduce that $$- \int^{\infty}_{0} 1 + z \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - \frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} \right) dz = \frac{\pi (4 \nu^2 - 1)}{8}.$$ Let $$I(\nu) = \int^{\infty}_{0} \log \frac{\pi x H^{(1)}_{\nu}(x) H^{(2)}_{\nu}(x)}{2} \cdot dx.$$ Integrating by parts, and being a little bit careful about what happens at $0$, and expressing the derivatives on Hankel functions in terms of Hankel functions of other arguments, we find that $$I(\nu) = - \int^{\infty}_{0} 1 + x \left(\frac{H^{(1)}_{\nu-1}(x)}{ H^{(1)}_{\nu}(x)} - \frac{H^{(2)}_{\nu+1}(x)}{ H^{(2)}_{\nu}(x)} \right) dx = \frac{\pi (4 \nu^2 - 1)}{8}.$$ As noted in the comments, this was the identity to be proved. Alternatively, integration by parts also shows that $$\frac{\pi (\mu^2 - \nu^2)}{2} = I(\mu) - I(\nu) = \int^{\infty}_{0} \log \frac{H^{(1)}_{\mu}(x) H^{(2)}_{\mu}(x)}{H^{(1)}_{\nu}(x) H^{(2)}_{\nu}(x)} \cdot dx,$$ and hence $$\int^{\infty}_{0} \log \frac{J_{\mu}(x)^2 + Y_{\mu}(x)^2}{J_{\nu}(x)^2 + Y_{\nu}(x)^2} \cdot dx = \frac{\pi (\mu^2 - \nu^2)}{2}.$$

Edit: Proof that $H^{(1)}_{\nu}(z)$ has no zeros in the upper half plane for real $\nu > 0$.

I realized that the proof can be completed in a similar manner. Let $$G^{(1)}_{\nu}(z) = d \log H^{(1)}_{\nu}(z) =\frac{1}{2} \left(A^{(1)}_{\nu}(z) - B^{(1)}_{\nu}(z)\right).$$ Then we have an asymptotic formula, as before: $$A^{(1)}_{\nu}(z) \sim i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z} + O(z^{-3}),$$ $$B^{(1)}_{\nu}(z) \sim - i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) + \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z} + O(z^{-3}),$$ and thus $$G^{(1)}_{\nu}(z) \sim i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} + \ldots \right) - \frac{1}{2z} + O(z^{-3}),$$ This is valid for $z$ with argument in $[-\pi + \epsilon,\pi]$, so in particular is valid in the upper half plane. If $C_R$ denotes the circle in the upper half plane, we find that: $$\lim_{R \rightarrow \infty} \int_{C_R} (G^{(1)}_{\nu}(z) - i) dz = - \frac{\pi i}{2},$$ because it is $O(R^{-1})$ plus the contribution from the $1/(2z)$ term. Let $\Omega_R$ denote the boundary of the region bounded by by the interval $[-R,R]$ and $C_R$. The function $H^{(1)}_{\nu}(z)$ and thus $H^{(1)}_{\nu}(z) e^{-iz}$ is holomorphic in $\Omega_R$ away from $z = 0$. At zero (and $\nu > 0$) we have an asymptotic $$H^{(1)}_{\nu}(z) \sim - i \cdot \frac{\Gamma(\nu)}{\pi} \left(\frac{2}{z}\right)^{\nu}.$$

It follows that the number of zeros in the upper half plane is given, accounting for the singularity at $0$ (modified by a factor of two since this integral only accounts for half of the singularity) by $$\frac{\nu}{2} + \lim_{R \rightarrow \infty} \frac{1}{2 \pi i} \oint_{\Omega_R} d \log (H^{(1)}_{\nu}(z) e^{-iz}) dz$$ $$= \frac{\nu}{2} + \frac{1}{2 \pi i} \left( \lim_{R \rightarrow \infty} \int_{C_R} (G^{(1)}_{\nu}(z) - i) dz + \lim_{R \rightarrow \infty} \int_{-R}^{R} (G^{(1)}_{\nu}(z) - i) dz\right).$$ We computed the first integral above, hence it suffices to compute: $$\frac{\nu}{2} -\frac{1}{4} + \frac{1}{2 \pi i} \lim_{R \rightarrow \infty} \int_{-R}^{R} (G^{(1)}_{\nu}(z) - i) dz.$$ We may write this as the integral $$\frac{\nu}{2} -\frac{1}{4} + \frac{1}{2 \pi i} \int_{0}^{\infty} \left( G^{(1)}_{\nu}(z) + G^{(1)}_{\nu}(-z) - 2 i \right) dz.$$ This expression evaluates to an integer, which is the number of zeros of $H^{(1)}_{\nu}(z)$ in the upper half plane. In particular, evaluating this integral numerically for a random value (say $\nu = \pi$) shows that, for this value, it is equal to zero. Now suppose that we vary $\nu$. Since this integral evaluates to an integer, to complete the proof, it suffices to show that it varies continuously. For the integral over $[R,\infty)$ this is clear for large enough $R$ from the asymptotic formula. For small values, it suffices to show that $H^{(1)}_{\nu}(z)$ doesn't have any zeros on the real line $[0,R]$, since otherwise the integrand is continuous in $\nu$, and the continuity of the integral is clear. Now on the real axis, we have, by definition, $$H^{(1)}_{\nu}(z) = J_{\nu}(z) + i \cdot Y_{\nu}(z).$$ Since $\nu$ is real, it suffices to show that $J_{\nu}(z)$ and $Y_{\nu}(z)$ do not have any simultaneous zeros. However, the zeros of these Bessel functions are well known to interlace (see Watson, A treatise on the theory of Bessel functions), and the result is established. Edit: Actually, there's a much easier proof that $J_{\nu}(z)$ and $Y_{\nu}(z)$ do not have any common zeros --- they are linearly independent solutions to a second order ODE!

• I take my hat off (+1) – Start wearing purple May 27 '13 at 22:23
• I'm sure what I'm seeing is a glimpse of a virtuoso. (+1) – Sangchul Lee May 28 '13 at 5:21