# Is this integral solvable analytically?

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

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

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