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I are trying to solve the following definite integral $$ \int_{0}^{\infty}\exp\left(-x+\frac{x^{1-\gamma}}{1-\gamma}\right) {\rm d}x \quad\mbox{where}\ \gamma > 1\ \mbox{is a parameter}. $$ My derivation is \begin{align} \int^\infty_0 e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x =& - \int^\infty_0 e^{\frac{x^{1-\gamma}}{1-\gamma}} d e^{-x} \\ =& - e^{\frac{x^{1-\gamma}}{1-\gamma}} \cdot e^{-x} |^\infty_{x=0} + \int^\infty_0 e^{-x} d e^{\frac{x^{1-\gamma}}{1-\gamma}} \\ =& \int^\infty_0 x^{-\gamma} e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x \ . \end{align} I was stuck here. Is it possible to find an explicit solution for this integral?If yes, how can I proceed further? How about the case $0 < \gamma < 1$?

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  • $\begingroup$ For: $0<\gamma <1$ is:$$\sum _{j=0}^{\infty } \frac{(1-\gamma )^{-j} \Gamma (1+j (1-\gamma ))}{\Gamma (1+j)}$$ $\endgroup$ Jan 30, 2021 at 10:48
  • $\begingroup$ @MariuszIwaniuk. Thanks. How did you come up with this result? $\endgroup$
    – Stanley
    Jan 31, 2021 at 11:16
  • $\begingroup$ $$\exp \left(-x+\frac{x^{1-\gamma }}{1-\gamma }\right)=\exp (-x) \exp \left(\frac{x^{1-\gamma }}{1-\gamma }\right)=\exp (-x) \sum _{j=0}^{\infty } \frac{x^{j (1-\gamma )} (1-\gamma )^{-j}}{j!}=\sum _{j=0}^{\infty } \frac{\exp (-x) \left(x^{j (1-\gamma )} (1-\gamma )^{-j}\right)}{j!}$$ and you integrate expression inside sum. $\endgroup$ Feb 1, 2021 at 7:43

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This question makes me feeling younger (I need it) since, in my former research group, we studied the integrals for integer values of $\gamma$.

$$I_\gamma=\int^\infty_0 e^{-x+\frac{x^{1-\gamma}}{1-\gamma}} d x$$

$$I_2=2 K_1(2)$$ $$I_3=\frac{1}{\sqrt{\pi }}G_{0,3}^{3,0}\left(\frac{1}{2^3}| \begin{array}{c} 0,\frac{1}{2},1 \end{array} \right)$$ $$I_4=\frac{\sqrt{3}}{2 \pi }G_{0,4}^{4,0}\left(\frac{1}{3^4}| \begin{array}{c} 0,\frac{1}{3},\frac{2}{3},1 \end{array} \right)$$ $$I_5=\frac{1}{\sqrt{2} \pi ^{3/2}}G_{0,5}^{5,0}\left(\frac{1}{4^5}| \begin{array}{c} 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},1 \end{array} \right)$$ $$I_6=\frac{\sqrt{5}}{4 \pi ^2}G_{0,6}^{6,0}\left(\frac{1}{5^6}| \begin{array}{c} 0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1 \end{array} \right)$$ and you can notice simple patterns in the Meijer G functions.

For the non-integer (but rational) values of $\gamma > 1$, I checked using a CAS and the results still express in terms of Meijer G functions.

From a numerical point of view, for large values of $\gamma$, the product $\gamma I_\gamma$ is quite close to linearity.

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  • $\begingroup$ Thanks. Your answer is very helpful. Any hope of dealing with a general $\gamma$, which might be non-integer valued or even irrational? $\endgroup$
    – Stanley
    Jan 30, 2021 at 6:12
  • $\begingroup$ @Stanley. My hope is $\ll \epsilon$ $\endgroup$ Jan 30, 2021 at 6:13

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