I am trying to figure out how the integral identities $$ \int_0^\infty J_0\left(2a\sinh\left(\frac{x}{2}\right)\right)\sin\left(bx\right)dx = \frac{2}{\pi}\sinh\left(\pi b\right)\left[K_{ib}\left(a\right)\right]^2 $$ and $$ \int_0^\infty Y_0\left(2a\sinh\left(\frac{x}{2}\right)\right)\cos\left(bx\right)dx = -\frac{2}{\pi}\cosh\left(\pi b\right)\left[K_{ib}\left(a\right)\right]^2 $$ are derived. (under the assumptions $a>0$, $b>0$)

Focusing for now on the first one, and guided by the symmetry between the expressions, I tried to expand the integrals on the left using the Mehler-Sonine integral representation, getting (for the first identity) $$ \frac{2}{\pi}\int_0^\infty\int_0^\infty \sin\left(2a\sinh\left(\frac{x}{2}\right)\cosh\left(t\right)\right)\sin\left(bx\right)dtdx $$ However that doesn't seem to get me much further as (I think?) these don't satisfy any condition allowing exchange of integrals.

Trying to work backwards from the answer the only real obvious expansion I could think of was $$ \frac{2}{\pi}\sinh\left(\pi b\right)\left[K_{ib}\left(a\right)\right]^2 = \frac{2}{\pi}\sinh\left(\pi b\right)\left[\frac{1}{\sin\left(\frac{1}{2}i\pi b\right)}\int_0^\infty\sin\left(a\sinh\left(t\right)\right)\sinh\left(ibt\right)dt\right]^2\\ =\frac{2}{\pi}\sinh\left(\pi b\right)\left[\frac{1}{\sinh\left(\frac{1}{2}\pi b\right)}\int_0^\infty\sin\left(a\sinh\left(t\right)\right)\sin\left(bt\right)dt\right]^2 $$ But I don't really see a way to cleverly combine the two integrals here to get back to the expression above either. What is a viable approach to derive these identities?


I am trying to compute variants of the above integrals which I cannot find in books with tables. In particular, I am trying to compute $$ \lim_{T\rightarrow\infty}\frac{1}{T}\int_0^TxJ_0\left(2a\sinh\left(\frac{x}{2}\right)\right)\sin\left(bx\right)dx $$ and $$ \lim_{T\rightarrow\infty}\frac{1}{T}\int_0^TxY_0\left(2a\sinh\left(\frac{x}{2}\right)\right)\cos\left(bx\right)dx $$ again assuming $a>0$ and $b>0$. My hope is that the techniques for the above somewhat standard integrals perhaps also apply to these, or at least can offer some inspiration for how to approach these.

  • $\begingroup$ Did we answer your question? $\endgroup$
    – Gary
    Mar 5 at 3:23
  • $\begingroup$ I'm gonna need a bit of time to work through both of the answers, but I think in principle yes. I'll mark them as the answer once I have done so. However, I may eventually come with the followup question of how to derive 10.32.16 if I can't figure that out on my own. $\endgroup$
    – davidv1992
    Mar 6 at 14:25
  • $\begingroup$ This paper could be of interest for you. $\endgroup$
    – Gary
    Mar 6 at 23:47

2 Answers 2


The following answer shows that $$\small \left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \, \cosh(\pi b) \int_{0}^{\infty} Y_{0}\left(2 a \sinh \frac{t}{2} \right) \cos(bt) \, \mathrm dt + \frac{\pi}{2} \, \sinh(\pi b) \int_{0}^{\infty} J_{0}\left(2 a \sinh \frac{t}{2} \right) \sin(bt) \, \mathrm dt =0. $$

The two integral identities satisfy this equation since $- \cosh^{2}(\pi b)+ \sinh^{2}(\pi b) = -1$.

This equation can be used with Gary's answer to prove the second integral identity.

My starting point is the more well known integral representation $$\left[ K_{\nu}(x) \right]^{2} = 2 \int_{0}^{\infty}K_{0}(2x \cosh t) \cosh(2 \nu t) \, \mathrm dt , \, \quad x>0. $$

Using this representation, we have $$\left[ K_{ib}(a) \right]^{2} = 2 \int_{0}^{\infty} K_{0} (2a \cosh x) \cos(2bx) \, \mathrm dx = \int_{0}^{\infty} K_{0} \left(2a \cosh \frac{u}{2} \right) \cos(bu) \, \mathrm du.$$

The important thing to notice here is that $\left[ K_{ib}(a) \right]^{2}$ is a real value if $a$ and $b$ are positive.

The function $$ f(z) = K_{0} \left(2a \cosh \frac{z}{2} \right) \cos(bz) $$ has branch cuts were $\cosh \left(\frac{x}{2} \right)$ is real and negative.

None of these branch cuts fall inside or on a rectangular contour with vertices at $z=0$, $z= R$, $z= R +\pi i $, and $z= \pi i $.

There is, however, a branch point on the contour at $z= i \pi$.

So let's integrate $f(z)$ counterclockwise around the above contour with the addition of an quarter circle indentation about $z = i \pi$.

However, since $ \lim_{z \to i \pi} (z- i \pi)K_{0} \left(2a \cosh \frac{z}{2} \right) \cos(bz) =0$, there is no contribution from letting the radius of the indentation go to zero. (The function $K_{0}\left( 2a \cosh \frac{z}{2} \right)$ behaves like $-\log(z-i \pi)$ near $z= i \pi$.)

As $R \to \infty$, the integral vanishes on the right side of contour because the magnitude of $K_{0}(z)$ decays exponentially to zero as $\Re(z) \to + \infty$.

On the left side of the contour, we have $$-i \int_{0}^{\pi} K_{0} \left(2a \cos \frac{t}{2} \right) \cosh (bt) \, \mathrm dt. $$

And on the top side of the contour, we have $$ \begin{align} &-\int_{0}^{\infty} K_{0} \left (2a \cosh \frac{t+ i \pi}{2} \right) \cos \left(b(t + i \pi) \right) \, \mathrm dt \\ &= - \int_{0}^{\infty} K_{0} \left(2ai \sinh \frac{t}{2} \right) \left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt \\ & \overset{(1)}{=} \frac{\pi}{2} \int_{0}^{\infty}\left(Y_{0} \left(2a \sinh \frac{t}{2} \right) + i J_{0} \left(2a \sinh \frac{t}{2} \right) \right)\left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt. \end{align}$$

Therefore, since there are no singularities inside the contour, we have $$\left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \int_{0}^{\infty}\left(Y_{0} \left(2a \sinh \frac{t}{2} \right) + i J_{0} \left(2a \sinh \frac{t}{2} \right) \right)\left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt$$

$$-i \int_{0}^{\pi} K_{0} \left(2a \cos \frac{t}{2} \right) \cosh (bt) \, \mathrm dt =0. $$

And by equating the real parts on both sides of the above equation, we get $$ \small \left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \, \cosh(\pi b) \int_{0}^{\infty} Y_{0}\left(2 a \sinh \frac{t}{2} \right) \cos(bt) \, \mathrm dt + \frac{\pi}{2} \, \sinh(\pi b) \int_{0}^{\infty} J_{0}\left(2 a \sinh \frac{t}{2} \right) \sin(bt) \, \mathrm dt =0. $$

$(1)$ See the answer here.

  • 1
    $\begingroup$ I completed your argument. $\endgroup$
    – Gary
    Mar 2 at 4:51

I shall prove the first identity. If you combine this with the other answer, the second identity should follow. By $(10.32.16)$ we have $$ I_{{\rm i}b} (a)K_{{\rm i}b} (a) = \int_0^{ + \infty } {J_0 (2a\sinh t){\rm e}^{ - 2{\rm i}bt} {\rm d}t} = \frac{1}{2}\int_0^{ + \infty } {J_0\! \left( {2a\sinh \left( {\frac{x}{2}} \right)} \right)\!{\rm e}^{ - {\rm i}bx} {\rm d}x} .$$ Similarly, $$ I_{ - {\rm i}b} (a)K_{{\rm i}b} (a) = I_{ - {\rm i}b} (a)K_{ - {\rm i}b} (a) = \frac{1}{2}\int_0^{ + \infty } {J_0\! \left( {2a\sinh \left( {\frac{x}{2}} \right)} \right)\!{\rm e}^{{\rm i}bx} {\rm d}x} . $$ Thus $$ - {\rm i}(I_{ - {\rm i}b} (a) - I_{{\rm i}b} (a))K_{{\rm i}b} (a) = \int_0^{ + \infty } {J_0\! \left( {2a\sinh \left( {\frac{x}{2}} \right)} \right)\!\sin (bx){\rm d}x} . $$ Then, by $(10.27.2)$, $$ \frac{2}{\pi }\sinh (\pi b)\left[ {K_{{\rm i}b} (a)} \right]^2 = \int_0^{ + \infty } {J_0\! \left( {2a\sinh \left( {\frac{x}{2}} \right)} \right)\!\sin (bx){\rm d}x} $$ which is the desired equality.


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