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I need to find the solution of this integral:

$$\int^\pi_{-\pi}{d\varphi\,{}\frac{\Gamma(n,2\pi\varphi)}{(1+ia\varphi)^n}},\tag{1}\label{1}$$

where $a\in(0,1)$ and $n$ is a positive integer (not zero). The solution should take a form as closed as possible, i.e., not contain a sum or another integral. Since neither Mathematica nor Maple offers a direct solution, I've been able to reduce the problem to solving either of the two following expressions:

\begin{gather*} i\int_{-2\pi}^{2\pi}dt\,{}e^{-t}\left(\frac{1}{t}+\frac{ia}{2}\right)^{1-n},\tag{2}\label{2} \\ i^{n+1}\sum_{k=0}^\infty\frac{(-a)^k\Gamma(n+k)}{2^k\Gamma(k+2)}\left[\Gamma(n+k+1,2i\pi)-\Gamma(n+k+1,-2i\pi)\right].\tag{3}\label{3} \end{gather*}

All three expressions are real and numerically can be seen to converge. The gamma function with two arguments is the incomplete gamma function, defined as

$$\Gamma(a,z)\equiv\int_{z}^\infty dt\,{} t^{a-1}e^{-t}.$$

A solution to any of \eqref{1}, \eqref{2}, or \eqref{3} depending only on $a$ and $n$ would be absolutely great, as well as a definitive argument on why they cannot be written in terms of the known functions (elementary, hypergeometric, etc).

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$$I_n(a)=\int^\pi_{-\pi}{d\varphi\,{}\frac{\Gamma(n,2\pi\varphi)}{(1+ia\varphi)^n}}.$$

For general $n$ I have a closed form expression for $a=0$, $$I_n(0)=-(2\pi n)^{-1}\left[e^{-2 \pi ^2} (2\pi^2)^{n+1} \left(1+e^{\pi (4 \pi +i n)}\right)+\left(2 \pi ^2+n\right) \Gamma \left(n+1,-2 \pi ^2\right)+\left(2 \pi ^2-n\right) \Gamma \left(n+1,2 \pi ^2\right)\right].$$

For $a\neq 0$ I only have closed form expressions for definite $n$, in terms of exponential integral functions: $$I_1(a)=\frac{1}{a}i e^{-2 i \pi /a} \left[\text{Ei}\bigl(2 \pi (i/a+\pi )\bigr)-\text{Ei}\bigl(-2 \pi ( \pi -i/a)\bigr)\right],$$ $$I_2(a)=\frac{4\pi}{a^3} \left[\frac{a \left(\sinh \left(2 \pi ^2\right)+\pi ^2 a \left(\pi (\pi a-i) E_{-1}\left(-2 \pi ^2\right)+\pi (\pi a+i) E_{-1}\left(2 \pi ^2\right)+a \sinh \left(2 \pi ^2\right)\right)\right)}{\pi ^2 a^2+1}+i \pi e^{-\frac{2 i \pi }{a}} \left(\text{Ei}\left(-\frac{2 \pi (a \pi -i)}{a}\right)-\text{Ei}\left(\frac{2 \pi (\pi a+i)}{a}\right)\right)\right],$$ $$I_3(a)=a^{-5} \left(\pi ^2 a^2+1\right)^{-2}e^{-\frac{2 \pi (\pi a+2 i)}{a}} \left[8 i \pi ^3 e^{\frac{2 \pi (\pi a+i)}{a}} (\pi +i a) \left(\pi ^2 a^2+1\right)^2 \left(\text{Ei}\bigl(2 \pi \left(i/a+\pi \right)\bigr)-\text{Ei}\bigl(-2 \pi ( \pi -i/a)\right)\bigr)+a e^{\frac{4 \pi (\pi a+i)}{a}} \left(i a^3-2 i \pi ^2 \left(a^2-1\right) a+4 \pi ^3 \left(a^2+1\right)-4 i \pi ^4 a\right) (\pi a-i)^2-i a e^{\frac{4 i \pi }{a}} (\pi a+i)^2 \left(a^3+2 \pi ^2 \left(a^2+1\right) a+4 i \pi ^3 \left(a^2-1\right)+4 \pi ^4 a\right)\right],$$ and progressively more complicated expressions for larger $n$.

Notation: $E_n(x)=\int_1^\infty e^{-xt}t^{-n}\,dt$, $\text{Ei}(x)=-\int_{-x}^\infty e^{-t}t^{-1}\,dt$.

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  • $\begingroup$ I know this, but I was hoping to get an expression for arbitrary $n$, or at least two different expressions for odd and even $n$, or something similar. Thank you, though. $\endgroup$ Mar 21, 2022 at 15:15

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