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I'm looking for an analytical solution of the following two integrals

$$\int_0^{\infty } \frac{x^{9/2}}{\left(A+x^3\right) \left(B+e^x\right)} \, \mathrm dx$$

and

$$\int_0^{\infty } \frac{x^3}{\left(A+x^3\right) \left(B+e^x\right)} \, \mathrm dx$$

with

$$A,B\in\mathbb{R} \land A,B\geq0$$

Wolfram Alpha gives up unfortunately. For

$B=0$

solutions exist.

These integrals have a physics background. They result from the Cronwell-Weisskopf approximation of calculating the energy averaged doping scattering times to get the dielectric function of a doped semiconductor. The usual theory is based on non-degenerate statistics, whereas I'm working on an implementation using degenerate statistics (needed for high doping concentrations). Within this framework the above integrals occur.

Rainer
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  • $\int_0^\infty x^2/((1+x^3)(1+e^x));dx \approx 0.30690840544564$ is not known to the ISC http://isc.carma.newcastle.edu.au/ – GEdgar Mar 27 '13 at 20:15
  • @ron, sorry corrected the wrong tagging – Rainer Mar 27 '13 at 20:32
  • They are not indefinite – Seyhmus Güngören Mar 27 '13 at 21:28
  • corrected text, they are of course definite.. – Rainer Jun 02 '13 at 08:52
  • Why this family of integrals? Where do they come from? – Did Jun 02 '13 at 14:19
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    These integrals have a physics background. They result from the Cronwell-Weisskopf approximation of calculating the energy averaged doping scattering times to get the dielectric function of a doped semiconductor. The usual theory is based on non-degenerate statistics, whereas I'm working on an implementation using degenerate statistics (needed for high doping concentrations). Within this framework the above integrals occur. – Rainer Jun 02 '13 at 14:47
  • @Rainer I would encourage editing your question to include that information - it may make the question more likely to be found by someone with knowledge of those approximations. – Steven Stadnicki Oct 27 '13 at 19:38

2 Answers2

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\begin{align*} \int_{0}^{\infty} \frac{x^3}{(A+x^{3})(B+e^{x})} \, dx &= \int_{0}^{\infty} \frac{A+x^{3}-A}{(A+x^{3})(B+e^{x})} \, dx \\ &= \int_{0}^{\infty} \frac{dx}{B+e^{x}} - \int_{0}^{\infty} \frac{A}{(A+x^{3})(B+e^{x})} \,dx \end{align*} Same trick can be applied for first integral.

Steven Stadnicki
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user103627
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  • Welcome to Math.SE! Here is a tip: Use of $\LaTeX$ syntax is so highly recommended in this forum that is is now considered as a necessity. – Sangchul Lee Oct 27 '13 at 18:55
  • @sos440 I think your edit is not correct. It should be fraction in first integral I suppose. – Cortizol Oct 27 '13 at 19:25
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Applying the expansion $$ \left( {{\rm e}^{x}}+B \right) ^{-1}= \sum _{k=0}^{\infty }{\frac { \left( -{{\rm e}^{-x}} \right) ^{k}{B}^{ k}}{{{\rm e}^{x}}}}, B>0,B<1,x \ge 0, $$ the integral $$\int_0^{\infty } \frac{x^3}{\left(A+x^3\right) \left(B+e^x\right)} \, \mathrm dx $$ is reduced to the series of the integrals $$\int_0^\infty \frac {(-1)^kx^3 e^{-\left( k+1 \right) x}B^k} {A+x^3}\,dx$$ which are calculated by Maple in a closed form. See the ouput here.

PS. The same with $x^{9/2}$ instead of $x^3 .$

user64494
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