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I have following integral $$\int_0^{\infty}e^{-ax-bx^m}dx$$ where $a>0, b>0, m>1$. I can get an approximation for the above integral when $b$ is small. However, I want to get an expression for the case when $b$ is large. Any help in this regard will be highly appreciated. Thanks in advance.

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  • $\begingroup$ Can't you do a u-substitution to get to the case of $b$ small. $\endgroup$ – mathworker21 Jan 24 '18 at 8:15
  • $\begingroup$ @mathworker21 I am not interested in the case when $b$ is small. I am interested in the case when $b$ is not small. $\endgroup$ – Frank Moses Jan 24 '18 at 8:18
  • $\begingroup$ You said you know what to do when $b$ is small. I'm saying, if $b$ is large, then do a u-substitution to get $b$ to be small, and then apply the approximation you know holds. $\endgroup$ – mathworker21 Jan 24 '18 at 8:21
  • $\begingroup$ @mathworker21 means $x=s^{-1}u$ where $s$ is very large value. Is this what you mean? $\endgroup$ – Frank Moses Jan 24 '18 at 8:26
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    $\begingroup$ @FrankMoses: in such a case it is pretty accurate, and a lower bound can be designed by applying a step of integration by parts, then the Cauchy-Schwarz inequality. $\endgroup$ – Jack D'Aurizio Jan 25 '18 at 2:15
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Expanding the factor $e^{-ax}$ in series and integrating term-by-term yields

$$ \int_0^\infty e^{-bx^m} e^{-ax}\,dx = \frac{b^{-1/m}}{m} \sum_{k=0}^{\infty} \frac{(-a)^k \operatorname{\Gamma}\!\left(\frac{k+1}{m}\right)}{k!} b^{-k/m}. $$

This series also serves as an asymptotic series as $b \to +\infty$.

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  • $\begingroup$ Why asymptotic? Does it not converge? $\endgroup$ – Yuriy S Jan 24 '18 at 9:26
  • $\begingroup$ @YuriyS it converges and is an asymptotic expansion. $\endgroup$ – Antonio Vargas Jan 24 '18 at 19:36
  • $\begingroup$ Thank you for clarification $\endgroup$ – Yuriy S Jan 24 '18 at 19:37

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