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I'm having trouble computing an integral.

$$ I=\int_0^1 \frac{\mathrm{d}x}{2x(1-x)}\left(x-\cosh\left(\frac{t\sqrt{1-x}}{\tau}\right)+\sqrt{1-x}\text{ }\mathrm{sinh}\left(\frac{t\sqrt{1-x}}{\tau}\right)\right)\left(1-\cos\left(\frac{y\sqrt{x}}{2u\tau}\right)\right) $$

I tried several changes of variable, such as $x=\cos^2(\frac{\theta}{2})$:

$$ I=\int_0^\pi \frac{\mathrm{d}\theta}{\sin(\theta)} \left( \underbrace{\cos^2(\theta)-1}_{A_1}+ \underbrace{1-\cosh\left( \frac{t}{\tau}\sin(\theta/2) \right)}_{A_2}+ \underbrace{\sin\frac{\theta}{2} \text{ }\mathrm{sinh}\left( \frac{t}{\tau}\sin(\theta/2) \right)}_{A_3} \right) \left( 1- \cos\left( \frac{y}{2u\tau}\cos(\theta/2)\right) \right) $$

The first term $A_1$ is computable, result is something with cosintegral, and is not a problem. One can consider $\frac{\mathrm d A_3}{\mathrm d y}$ and one gets: $$ \frac{\mathrm d A_3}{\mathrm d y}=\frac{1}{4u\tau} \int_0^\pi\mathrm d \theta \text{ } \mathrm{sinh}\left( \frac{t}{\tau}\sin(\theta/2) \right) \sin\left( \frac{y}{2u\tau}\cos(\theta/2)\right) $$

The $A_2$ term can be transform in something very similar: $$ \frac{\mathrm d^2 A_3}{\mathrm d t\mathrm d y}=- \frac{1}{4u\tau^2} \underline{\int_0^\pi\mathrm d \theta \text{ } \mathrm{sinh}\left( \frac{t}{\tau}\sin(\theta/2) \right) \sin\left( \frac{y}{2u\tau}\cos(\theta/2)\right)}_{\Large\text{this one}} $$

Has anyone any idea how to compute such an integral ? I tried looking in the handbook for mathematical functions, and found that integral representation of Kelvin functions are almost the same as my integral. I am looking for any special (or not) function that may allow me to have an analytic expression of the latter integral.

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  • $\begingroup$ What if.... there is no way of doing so? I mean from the looks of it I don't think there will be any closed form. Perhaps I'm wrong though. Did you Wolfram it? $\endgroup$ – user88595 Mar 27 '14 at 10:35
  • $\begingroup$ Yes I did, Mathematica doesn't give a closed form. There might be no way to compute the integral, I was just asking the question hoping somebody might have a brilliant idea. I'd live with it uncomputed. $\endgroup$ – ChocoPouce Mar 27 '14 at 10:50
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One way to get an expression for the integral, although not in closed form, is to use the Jacobi-Anger expansion: $${{\rm e}^{iz\cos \left( \theta \right) }}=\sum _{n=-\infty }^{\infty } {i}^{n}{{\rm J}_n\left(z\right)}{{\rm e}^{in\theta}} \tag{1}$$ to expand the integrand in terms of Bessel functions of the first kind ($J_n$). In so doing, it can be shown that: $$\sinh \left( {\frac {t\sin \left( \frac{\theta}{2} \right) }{\tau}} \right) \sin \left( {\frac {y\cos \left( \frac{\theta}{2} \right) }{2u\tau}} \right) =\sum_{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty} i\left( -1 \right) ^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)} {{\rm J}_{2n+1}\left({\frac {-it}{\tau}}\right)}\sin \left( \frac{\, \left( 2\,n+1 \right) \theta}{2} \right) \cos \left( \frac{\, \left( 2\,k+1 \right) \theta}{2} \right) \right) \tag{2}$$ which can be written: $$\frac{i}{2}\sum _{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty}\left( -1 \right) ^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)} \left( {{\rm J}_{-2k-1+2n}\left({\frac {-it}{\tau}}\right)}+ {{\rm J}_{2k+1+2n}\left({\frac {-it}{\tau}}\right)} \right) \sin \left( n\theta \right) \right) \tag{3}$$ and thus:

$$\int _{0}^{\pi }\!\sinh \left( {\frac {t\sin \left( 1/2\,\theta \right) }{\tau}} \right) \sin \left( {\frac {y\cos \left( 1/2\,\theta \right) }{2u\tau}} \right) {d\theta}=\sum _{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty}\frac{i\left( -1 \right) ^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)} \left( {{\rm J}_{-2k+1+4n}\left({\frac {-it}{\tau}}\right)}+ {{\rm J}_{2k+3+4n}\left({\frac {-it}{\tau}}\right)} \right) }{2n+1} \right) \tag{4}$$

It may not look too pretty but the series often converges very rapidly and it may simplify further. Convolutions of Bessel functions are sometimes known as generalised Bessel functions although this may be a stretch in terms of analytic functions.

Update

This way is probably neater and simpler. Note that:

$$\sinh \left( {\frac {t\sin \left( \frac{\theta}{2} \right) }{\tau}} \right) \sin \left( {\frac {y\cos \left( \frac{\theta}{2} \right) }{2u \tau}} \right) =-\frac{i}{2} \left( \cos \left( x\sin \left( \frac{\theta}{2}+ \phi \right) \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi \right) \right) \right) \tag{i}$$ $$\phi=i\mathrm{arctanh} \left( {\frac {y}{2tu}} \right) \in \mathbb{C},\quad x=\frac{it}{2\tau}{\sqrt{4-{\frac {{y}^{2}}{{t}^{2}{u}^{2}}}}} \in \mathbb{C} \tag{ii} $$

where standard product-to-sum trig identities were used and then the two trig functions inside, each with the same argument, are written as one scaled and shifted trig function.

Then starting from $(1)$ you can prove:

$$-\frac{i}{2}\left[\cos \left( x\sin \left( \frac{\theta}{2}+ \phi \right) \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi \right) \right)\right] =\\ i\sum _{n=-\infty }^{\infty }{{\rm J}_n\left(x\right)}\sin \left( n\frac{\theta}{2}\right) \sin \left( n\phi \right) \tag{iii}$$

integrate and simplify to get:

$$\int _{0}^{\pi }\!-\frac{i}{2}\left[\cos \left( x\sin \left( \frac{\theta}{2}+ \phi \right) \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi \right) \right)\right] {d\theta}=\\4\,i\sum _{n=1}^{\infty }{\frac { {{\rm J}_{4n-2}\left(x\right)}\sin \left( \left( 4n-2 \right)\, \phi \right) }{2\,n-1}} \tag{iv}$$

which checks out for all numerical values I have tried so far. Note that the integrals on the left individually look like Bessel functions $(J_0)$ but for the $\phi$ shift.

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  • $\begingroup$ Thanks for this great answer. I'm gonna accept it as soon as I'd get all the steps. I have an issue with (2). I think the sums should start at $n=0$ and $k=0$. I also have a $-4$ prefactor in (2). $\endgroup$ – ChocoPouce Mar 27 '14 at 16:40
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    $\begingroup$ I have checked $(2)$ it seems correct to me. When testing the sums you can run them from $-N..N$ and it converges quite quickly. I like to start from the symmetric sums as it is easier to shift indices around without them falling off the bottom. I have added something that may be easier although it uses complex variables. $\endgroup$ – Graham Hesketh Mar 27 '14 at 18:14
  • $\begingroup$ (iv) is impressive. It saddens me though because I'm not sure it can be integrated over $y$ or $t$ (in order to get the first integral in my question). (4) is a better form to do so, but I will end up with 4 sums, and evaluating the integral numerically might be as fast as the evaluating the sums. $\endgroup$ – ChocoPouce Mar 28 '14 at 15:04
  • $\begingroup$ I have trouble getting $(3)$. Is there a formula to go from $(2)$ to $(3)$? I tried expanding the sine-cosine product, but I don't see how one can end up with both $k$ and $n$ indices in Bessel functions in $(3)$. $\endgroup$ – ChocoPouce Apr 15 '14 at 10:04
  • $\begingroup$ Write: $\sin \left( \frac{1}{2} \left( 2\,n+1 \right) \theta \right) \cos \left( \frac{1}{2} \, \left( 2\,k+1 \right) \theta \right) =\frac{1}{2}\sin \left( \theta\, \left( n+1+k \right) \right) +\frac{1}{2}\sin \left( \theta\, \left( n-k \right) \right)$ then split the sum over $n$ into two parts one $\sin$ term in each. In one sum, you change the $n$ index by writing $n\rightarrow n-1-k$ and in the other you write $n\rightarrow n+k$. For piece of mind you might imagine the $n$ sum inside the $k$ sum but it doesn't matter because the sums are infinite. You then recombine the sums. $\endgroup$ – Graham Hesketh Apr 15 '14 at 10:32

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