Fix $a>0$ and $b>0$. Then Gradshteyn and Ryzhik give the following integrals 6.796.4 and 6.796.5: $$ \int_0^\infty \cos(bx) \cosh(\pi x) \left[ K_{ix}(a) \right]^2 \,dx = - \frac{\pi^2}{4} Y_{0}\left( 2 a \sinh\left( \tfrac{b}{2} \right) \right) \\ \int_0^\infty \sin(bx) \sinh(\pi x) \left[ K_{ix}(a) \right]^2\, dx = \frac{\pi^2}{4} J_{0}\left( 2 a \sinh\left( \tfrac{b}{2} \right) \right) \\ $$

I am in the very unfortunate situation where I need a very similar integral which is not listed: $$ \int_0^\infty \cos(bx) \sinh(\pi x) \left[ K_{ix}(a) \right]^2 \,dx = ?? $$

I have tried playing around with the integrals given in G+R (mainly hitting the integrals with $\frac{d}{db}$) but I am not able to do anything useful.

Furthermore, G+R list Erdelyi's Integral Transforms Volume II as the source for these two integrals. But I've gone and checked this reference, and there is no derivation provided - these two integrals are simply listed there.

Is there any way to evaluate $\int_0^\infty \cos(bx) \sinh(\pi x) \left[ K_{ix}(a) \right]^2 \,dx?$

  • 1
    $\begingroup$ The second integral from G & R listed above makes no sense. On the LHS, when b=0 the integral is rightly zero. On the RHS however, the integral is $\pi^2/4$. Am I wrong? $\endgroup$
    – Ron Gordon
    May 29, 2018 at 15:56
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    $\begingroup$ @RonGordon. Integrals are correct for $a>0$ and $b>0$. $\endgroup$ May 29, 2018 at 16:04
  • $\begingroup$ Could it help to include a parameter inside of the hyperbolic trig function so that you can try a parametric differentiation trick? $\endgroup$ May 29, 2018 at 19:13

1 Answer 1


Here's a formal derivation of the second relationship, and a perturbation to give an alternative representation for the integral of question. Unfortunately it doesn't look much simpler.

Use Gradshteyn 6.664.1 with $\nu=i\,x$ to get $$ \big(K_{ix}(a)\big)^2 = \frac{\pi}{\sinh(\pi \, x)} \int_0^\infty J_0(2a\sinh(t))\,\sin(2xt) \, dt.$$

Insert this into the left-hand side and interchange integration to get

$$\int_0^\infty \sin(bx) \sinh(\pi\,x)\big(K_{ix}(a)\big)^2 \, dt = \pi \int_0^\infty \,dt \,J_0(2a\sinh(t))\, \int_0^\infty \sin(bx)\sin(2xt) \, dx $$ The product of sines becomes a sum of cosines, and the integral of the cosine is the Dirac delta function, up to a factor; i.e., $$ \int_0^\infty \sin(bx)\sin(2xt) \, dx = \frac{1}{2}\Big(\pi\, \delta(2t-b) - \pi \,\delta (2t+b) \Big).$$ Using the fact that $\delta(ax)=\delta(x)/a$, the fact that the second delta function is zero for b>0, and the shifting property of $\delta$ completes the proof of

$$\int_0^\infty \sin(bx) \sinh(\pi\,x)\big(K_{ix}(a)\big)^2 \, dt = \frac{\pi^2}{4} \,J_0(2a\sinh(b/2)). $$

For the integral of question, the same steps lead to the representation $$\int_0^\infty \cos(bx) \sinh(\pi\,x)\big(K_{ix}(a)\big)^2 \, dt = \pi \int_0^\infty \,dt \,J_0(2a\sinh(t))\, \int_0^\infty \cos(bx)\sin(2xt) \, dx .$$

However $\cos(bx)\sin(2xt) = 1/2(\sin(x(2t+b))+\sin(x(2t-b))$ and the integral over the positive real line of the sine is not a delta function with the shifting property: it is a Cauchy principal value integral. Thus the final result is

$$\int_0^\infty \cos(bx) \sinh(\pi\,x)\big(K_{ix}(a)\big)^2 \, dt= P.V. \frac{\pi}{2} \int_0^\infty J_0(2a\sinh(t))\,\frac{t\,dt}{t^2-(b/2)^2} .$$

This might be simpler if your numerics routines have to work hard to calculate $K_{ix}(a).$


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