# Solving defined integral

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:

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

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