Find Analytic Expression for Integral with Approximations

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SUMMARY

The discussion focuses on finding a closed form or analytic expression for the integral $$\int_{0}^{\infty} \frac{\frac{1}{x^4} e^{-\frac{r}{x^2}} e^{-\frac{r}{z^2}}}{\frac{1}{x^2} e^{-\frac{r}{x^2}} + \frac{1}{y^2} e^{-\frac{r}{y^2}}} dr$$, where \(x\), \(y\), and \(z\) are constants independent of \(r\). Participants are encouraged to share their thoughts on how to approach this problem. The integral involves exponential decay terms and rational functions, indicating a potential application of techniques such as Laplace transforms or residue calculus for evaluation.

PREREQUISITES
  • Understanding of integral calculus, particularly improper integrals.
  • Familiarity with exponential functions and their properties.
  • Knowledge of techniques such as Laplace transforms and residue calculus.
  • Experience with analytic expressions and closed-form solutions in mathematical analysis.
NEXT STEPS
  • Research methods for evaluating improper integrals, specifically those involving exponential decay.
  • Learn about Laplace transforms and their applications in solving integrals.
  • Study residue calculus and its use in complex analysis for evaluating integrals.
  • Explore examples of finding closed forms for integrals with similar structures.
USEFUL FOR

Mathematicians, students studying advanced calculus, and researchers in mathematical analysis seeking to understand integral evaluation techniques and closed-form solutions.

venkaiah
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Find the closed form (or) analytic expression form for the following integral

$$
\hspace{0.3cm} \large {\int_{0} ^{\infty} \frac{\frac{1}{x^4} \hspace{0.1cm} e^{- \frac{r}{x^2}}\hspace{0.1cm}e^{- \frac{r}{z^2}} }{ \frac{1}{x^2} \hspace{0.1cm} e^{- \frac{r}{x^2}}+ \frac{1}{y^2} \hspace{0.1cm} e^{- \frac{r}{y^2}}}} dr \hspace{.2cm} ; \hspace{1cm} x>0,y>0,z>0 $$ where $ x $ ,$ y $ and $z $ are constants and independent of $ r $.
 
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Hi venkaiah and welcome to MHB! :D

Any thoughts on how to begin?
 

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