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I have this integral that comes from my research with some Fourier Transforms of spectrum functions:

$$ G(\tau) = \int_{0}^{\infty} e^{-\Lambda x} x^n e^{i \tau ( c_1 - c_2 e^{-c_3 x} ) } dx $$

where $c_1, c_2, c_3, \Lambda, n > 0$.

The only way for me now is to use a series expansion of the term $e^{i \tau ( c_1 - c_2 e^{-c_3 x} ) }$ and take the first few terms. However, I was wondering if there is a smarter way to do it.

====== EDIT =========

Here, $i = \sqrt{-1}$.

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    $\begingroup$ Note $ \int e^{be^x} = \text{Ei}(be^x) + C$. Are $\Lambda, n$ integers? $\endgroup$ Commented Jul 6, 2023 at 12:45
  • $\begingroup$ @SidharthGhoshal no, they are not integers. $\endgroup$
    – CfourPiO
    Commented Jul 6, 2023 at 12:48

1 Answer 1

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there is a closed form solution for $$I_n = \int_{0}^{\infty} e^{-\Lambda x} x^n e^{i \tau ( c_1 - c_2 e^{-c_3 x} ) } dx$$ for integer $n$, for example, for $n=0$:

$$I_0=\frac{1}{\lambda (c_3+\lambda)}e^{i c_1 \tau}$$ $$\qquad\times \left[(c_3+\lambda) \, _1F_2\left(\frac{\lambda}{2 c_3};\tfrac{1}{2},\tfrac{\lambda}{2 c_3}+1;-\tfrac{1}{4} c_2^2 \tau^2\right)-i c_2 \lambda \tau \, _1F_2\left(\tfrac{\lambda}{2 c_3}+\tfrac{1}{2};\tfrac{3}{2},\tfrac{\lambda}{2 c_3}+\tfrac{3}{2};-\tfrac{1}{4} c_2^2 \tau^2\right)\right].$$

for larger integer $n$ the expressions are similar, in terms of hypergeometric functions, but much longer.

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  • $\begingroup$ Thank you very much for the response. How did you arrive at this answer? Did you use the Bessel function expansion of the term $e^{i \tau (c_1 - c_2 e^{-c_3 x})} $ ? $\endgroup$
    – CfourPiO
    Commented Jul 6, 2023 at 18:09
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    $\begingroup$ this is Mathematica output. $\endgroup$ Commented Jul 6, 2023 at 18:18

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