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I have the following integral: $$f(x) = \int_{1}^{x} \dfrac{t^{-\alpha}}{(x - t)^{\frac{3}{2}} } \exp{\left( \dfrac{-A^2 {t}^2}{(x - t)} \right)} dt, A > 0, \alpha > 0$$

I want to calculate the log-log slope of $f(x)$ for $x$ large, say $10^4$. I can see numerically (after numerical integration and plotting) that for certain range of values of $A$ and $\alpha$, the log-log slope (abs value) of $f(x)$ seems to scale as $1 + \alpha/2$, other times like $1 + \alpha$.

I would like to see if asymptotics can get me anywhere towards that, but this integral seems to not fit any of the standard forms like Laplace. Any ideas ?

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  • $\begingroup$ For large $x$ you are essentially averaging a function of a high variance Gaussian, except when $t$ gets close to $x$ (which I think negligibly contributes to the integration anyway). is there any way to change variables to make it a low variance Gaussian instead? $\endgroup$ – Ian Sep 7 '18 at 18:09
  • $\begingroup$ $f(x)\approx \frac{1}{x^{3/2} (-1+\alpha )}+\frac{-9+4 A^2+3 \alpha -2 A^2 \alpha }{2 x^{5/2} (-3+\alpha ) (-2+\alpha )}$ for: $A>>2$,$\alpha>>2$ and $x >> 1000$. $\endgroup$ – Mariusz Iwaniuk Sep 7 '18 at 20:50
  • $\begingroup$ @MariuszIwaniuk: Could you write that out in more detail in an answer ? $\endgroup$ – me10240 Sep 7 '18 at 21:17
  • $\begingroup$ I used Mathematica 11.3. Execute this code: n = 3;(*n Terms*)AsymptoticIntegrate[ t^-α/(x - t)^(3/2)*Exp[-A^2*t^2/(x - t)], {t, 1, Infinity}, {x, Infinity, n}, Assumptions -> {α > 0, A > 0}] $\endgroup$ – Mariusz Iwaniuk Sep 7 '18 at 21:29
  • $\begingroup$ Your integral can be approximated as a sum over confluent hypergeometric functions in the case I examined: $0<\alpha<1$ and large $A\,x.$ Section 13.8 of DLMF shows how there is a transition region as the parameters sweep below to much higher than a large argument. It would probably be helpful to narrow the scope of your problem. If you are O.K. with special functions as an answer, I can type it up over the weekend. $\endgroup$ – skbmoore Sep 7 '18 at 22:22
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For fixed $A$ and $\alpha$, the asymptotic can be found in a way similar to this; after the change of variables $\xi = t/(x - t)$, the leading term is determined by the maximum of the exponent at $\xi = 0$ but the integration range is $[1/(x-1), \infty)$: $$f(x) = \int_{1/(x - 1)}^\infty \frac 1 {\sqrt {x (\xi + 1)}} \left( \frac {x \xi} {\xi + 1} \right)^{-\alpha} e^{-A^2 x \xi^2 / (\xi + 1)} d\xi \sim \\ x^{-\alpha - 1/2} \int_{1/(x - 1)}^\infty \xi^{-\alpha} e^{-A^2 x \xi^2} d\xi = \\ \frac 1 2 A^{\alpha - 1} x^{-\alpha/2 - 1} \Gamma \!\left( \frac {1 - \alpha} 2, \frac {A^2 x} {(x - 1)^2} \right) \sim \\ \frac 1 2 A^{\alpha - 1} x^{-\alpha/2 - 1} \Gamma \!\left( \frac {1 - \alpha} 2, \frac {A^2} x \right) \sim \\ \cases { \frac 1 2 A^{\alpha - 1} \Gamma \!\left( \frac {1 - \alpha} 2 \right) x^{-\alpha/2 - 1} & $\alpha < 1$ \\ \frac 1 2 x^{-3/2} \ln x & $\alpha = 1$ \\ \frac 1 {\alpha - 1} x^{-3/2} & $\alpha > 1$ \\}, \quad x \to \infty.$$

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  • $\begingroup$ Could you explain the second step a bit more, about why the $\xi + 1$ term has been dropped from the integrand ? $\endgroup$ – me10240 Sep 9 '18 at 13:18
  • $\begingroup$ Also, I understand the derivation of the last part when $\alpha <1$ but It is not clear to me how you get the expressions when $\alpha \geq 1$. $\endgroup$ – me10240 Sep 9 '18 at 18:02
  • $\begingroup$ We're expanding the exponential and non-exponential parts around $\xi = 0$: $$\int_{1/(x - 1)}^\infty \phi(\xi) \xi^{-\alpha} e^{x f(\xi)} d \xi \sim \phi(0) e^{f(0) x} \int_{1/(x - 1)}^\infty \xi^{-\alpha} e^{f''(0) x \xi^2/2} d \xi.$$ For $a \leq 0$, the expansion of $\Gamma(a, z)$ around $z = 0$ is determined by the behavior of the integrand at $\xi = 0$: $$\int_z^\infty \xi^{a - 1} e^{-\xi} d \xi \sim \int_z^1 \xi^{a - 1} d \xi.$$ $\endgroup$ – Maxim Sep 9 '18 at 20:02
  • $\begingroup$ The expansion of $\phi$ and $f$ around zero---what is the name of this technique ? Is this the Laplace method ? I have been trying to get the general method from a book but not able to yet. $\endgroup$ – me10240 Sep 11 '18 at 1:18
  • $\begingroup$ The expansion is taken around zero only because the maximum of the exponent is at zero. For $\alpha < 1$, the integration range can be extended to $[0, \infty)$, then it's a standard case for Laplace's method (an extension of it to integrable singularities). The case $\alpha > 1$ can be done without Laplace's method as it's the exponential factor that can be replaced by a constant. $\alpha = 1$ is slightly more difficult. Bender, Orszag, Advanced Mathematical Methods for Scientists and Engineers has plenty of examples. Another good book is Erdelyi's Asymptotic expansions. $\endgroup$ – Maxim Sep 11 '18 at 7:18

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