5
$\begingroup$

I'm looking for a way of evaluating $$\int_0^\pi\sin x \exp(-a/\sin x)dx$$ to get a second order Bickley function $K_2(a)$, which is basically the same integral, but $\cos x$ instead of $\sin x$ and the limits change from $0$ to $\pi/2$, which is understandable.

I'm a bit lost what kind of variable substitution could I do. Any suggestions? I have tried $-a/\sin x = u$, but that doesnt seem to give reasonable results. Thanks in advance!

$\endgroup$
  • $\begingroup$ Why was this downvoted? Ashkhen has shown some effort. $\endgroup$ – Shaun Apr 16 '18 at 12:30
  • 1
    $\begingroup$ What are the integration limits? $\endgroup$ – Spine Feast Apr 16 '18 at 12:31
  • $\begingroup$ Sorry for not mentioning that; from 0 to π $\endgroup$ – Ashkhen Nalbandian Apr 16 '18 at 12:37
  • 2
    $\begingroup$ The function's symmetric over $x=\pi/2$ so you could do 2$\int_0^{\frac{\pi}{2}}\sin\left(x\right)\exp\left(\frac{-a}{\sin x}\right)dx$. This might be easier since the integrand is monotonic over $(0,\pi/2)$. $\endgroup$ – Jam Apr 16 '18 at 12:54
  • 1
    $\begingroup$ The integral is equal to $$f(a)=2\int_1^\infty e^{-a u} \frac{du}{u^2 \sqrt{u^2-1}}$$ Also $$f''(a)= 2 K_0(a)$$ where $K_0$ is the modified Bessel function of a second kind $\endgroup$ – Yuriy S Apr 16 '18 at 12:57
4
$\begingroup$

Elaborating on my comment.

Though I now consider the limits to be from $0$ to $\pi/2$, so the final function would have to be multiplied by $2$.

Substitution: $$u=\frac{1}{\sin x}$$

makes the integral:

$$f(a)=\int_1^\infty e^{-a u} \frac{du}{u^2 \sqrt{u^2-1}}=\int_0^\infty \frac{dt}{\cosh^2 t}e^{-a \cosh t}$$

This makes it obvious that:

$$f''(a)=K_0(a)$$

Where the initial conditions can be easily found from the original integral:

$$f(0)=1 \\ f'(0)=-\frac{\pi}{2}$$

Does $f(a)$ have a closed form? Probably not a nice one. We can always use numerical methods.

$\endgroup$
  • $\begingroup$ As you showed above, the integral results in modified bessel function of second order, which can be approximated by Bickley inetgral, and that can be further numerically solved in Matlab, for example. Thanks a lot, btw. $\endgroup$ – Ashkhen Nalbandian Apr 16 '18 at 13:30
  • $\begingroup$ PS In case you're interested; ams.org/journals/mcom/1978-32-143/S0025-5718-1978-0471262-6/… $\endgroup$ – Ashkhen Nalbandian Apr 16 '18 at 13:30
  • $\begingroup$ @AshkhenNalbandian, I have never heard of Bickley functions before now, but as they are defined by the integrals like that one, I don't really understand what do you want to do? Numerically evaluate the integral? Find some other closed form? Please clarify the question if there's something in particular you'd like to know $\endgroup$ – Yuriy S Apr 16 '18 at 13:33
  • $\begingroup$ At this point my questions is fully answered. I have to solve the Bickley function (mostly used in neutron transport theory, in reactor physics), but I just wanted to fill the gap (since it is not done in literature) of obtaining the Bickley function from the integral I posted. $\endgroup$ – Ashkhen Nalbandian Apr 16 '18 at 13:35
  • 1
    $\begingroup$ the first integral is just a combinations of struves and bessels, but the remaining one seems to be impossible $\endgroup$ – tired Apr 16 '18 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.