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I am trying to find a closed form expression for following definite integral

\begin{equation} \int_{1}^{\infty} e^{\frac{1}{2a}(\frac{Kb}{x}-x)} ~~x^{-\frac{K}{2}} ~dx \end{equation}

where $a,b$ are positive real numbers and $K$ is an positive integer. I have looked into almost all integral tables, I have also tried wolfram. An accurate approximation is also appreciated.

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  • $\begingroup$ There is no closed form solution but various asymptotic expansions are possible. What range of $a$, $b$ and $K$ are you interested in? $\endgroup$ – jcandy Nov 3 '17 at 18:31
  • $\begingroup$ @jcandy thank you for your comment, I appreciate if you can help me with approximates, a belongs roughly to [1,5] , and b roughly in [0.5,3], and K can be in [5, 100] and K is integer. $\endgroup$ – Alireza Nov 3 '17 at 23:36
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First, given your parameter ranges, I will rewrite your integral using new parameters $a$ and $b$ which are assumed to be $\cal{O}(1)$ and a (possibly) large parameter $\lambda$ (note that $\lambda$ is your $K/2$, etc): $$ I(a,b,\lambda) = \int_1^\infty dx \, e^{-ax} e^{\lambda \left(b/x-\ln x\right)} \; . $$ Crude approximation: First, we can write a simple crude approximation to the integral by noticing that the second exponential factor is a monotonicaly decreasing function of $x$. Then, writing $u=1+x$ and Taylor expanding the second exponential we find $$ \begin{eqnarray} I & \simeq & \int_0^\infty du \, e^{-a (1+u)} e^{\lambda b(1-u)} e^{-\lambda u} \; , \\ &=& e^{\lambda b -a} \, \frac{1}{a+\lambda b + \lambda} \; . \end{eqnarray} $$ As a point of comparison, for $I(1,1,3)$, the true value is $1.28$ and the approximation gives $1.06$.

Asymptotic expansion: For $\lambda \gg 1$, we can work out a correction to this formula. Perform the same variable change as before, but expand up to second order in $u$: $$ I \sim \int_0^\infty du \, e^{-a (1+u)} e^{\lambda b(1-u+u^2)} e^{-\lambda (u-u^2/2)} \; . $$ The integral is dominated by the region $u \sim 1/\lambda$, which means that the exponential of the quadratic terms can also be Taylor expanded to give $$ \begin{eqnarray} I &\sim& \int_0^\infty du \, e^{-a (1+u)} e^{\lambda b(1-u)} e^{-\lambda u} \left[ 1+ u^2(\lambda b + \lambda/2)\right] \\ &=& e^{\lambda b-a} \int_0^\infty du \, e^{-su} \left[ 1+ u^2(\lambda b + \lambda/2)\right] \; , \end{eqnarray} $$ where $s \doteq a + \lambda (b + 1)$. Performing the integrations gives $$ I(a,b,\lambda) \sim e^{\lambda b-a} \left( \frac{1}{s} + \frac{2 \lambda(b+1/2)}{s^3}\right) \; . $$ As a point of comparison, for $I(1,0.5,30)$, the true value is $2653.8$ and the asymptotic expansion gives $2650.5$, an improvement over the crude formula which gives $2564.0$. Actually, the asymptotic expansion seems to be pretty good even for smaller values of $\lambda$.

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