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Is there an analytical solution to the following integral:

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

Where:

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

And:

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

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.

Edit:

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:

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

And:

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

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  • $\begingroup$ 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
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I cannot do better than Mathematica !

enter image description here

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