Is there an analytical solution to the following integral:

$$ I = \iint\limits_{\mathcal{D}} \exp\left(-kx\right) \mathrm{d}x \mathrm{d}y $$


$$ \mathcal{D}(x,y) \equiv x^2 + y^2 \leq R^2 $$


$$ R, k \in \mathbb{R}^+_0$$

This integral arose in a simple problem (detector response to a cylindrical cell using Beer-Lambert law) but I am struggling to solve it. I did a polar coordinate change, but then I got:

$$ I = \iint\limits_{\mathcal{D}} \rho \exp \left( -k\rho \cos(\theta) \right) \mathrm{d}\rho \mathrm{d}\theta $$

Which seems not easier to solve.

I looked for this integral form in Handbook of Integrals but I cannot found it. I tried to solve it with a symbolic solver without success. I tried to apply the Green's Theorem but the mixed exponential/trigonometric term reappeared.


As pointed out by Santosh Linkha and JJacquelin (featuring Mathematica), this integral can be solved using First Kind modified Bessel function. Readers which are not used to Bessel functions - as I was, might be intersted to know that resolution of this integral requires the following identities:

$$ I_n(x) = \frac{1}{\pi}\int\limits_0^\pi \cos(n\theta)\exp(x\cos(\theta))\mathrm{d}\theta $$


$$ \frac{\mathrm{d}}{\mathrm{d}x} \left( x^\nu I_\nu(x) \right) = x^\nu I_{\nu-1}(x) $$

  • $\begingroup$ The result is $\frac{2 \pi R I_1(k R)}{k}$ from Mathematica $\endgroup$ – Santosh Linkha May 28 '14 at 12:50
  • $\begingroup$ And what is $I_1$? $\endgroup$ – jlandercy May 28 '14 at 12:54
  • $\begingroup$ Bessel function of first kind :(( $\endgroup$ – Santosh Linkha May 28 '14 at 12:55
  • $\begingroup$ Thanks for answering. Do you know how to get this result, I mean without calculator? $\endgroup$ – jlandercy May 28 '14 at 13:11
  • $\begingroup$ Errr ... looks like I could give a try $\endgroup$ – Santosh Linkha May 28 '14 at 13:12

I cannot do better than Mathematica !

enter image description here


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.