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The integral $$I=\int _{-\infty }^{+\infty }e^{-e^x+ax}dx \quad \textrm{ where }a>0 $$ seems very difficult. However, putting $y=e^x$ converts the integral into $$ I=\int_0^{+\infty} y^{a-1} e^{-y} d y=\Gamma(a) $$ Then I try to generalise it to $$ I_n=\int_{-\infty}^{+\infty} x^n e^{-e^x+a x} dx=\int_0^{+\infty} y^{a-1} e^{-y} \ln ^n y d y= \frac{\partial^n}{\partial a^n}(\Gamma(a)), $$ by which I am stuck and try some particular cases:

When $a=2$, $$\int_{-\infty}^{+\infty} xe^{-e^x+2x} dx =\Gamma^{\prime}(2)=\Gamma(2)\psi(2)= -\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) =-\gamma+1$$ $$ \begin{aligned} \int_{-\infty}^{+\infty} x^2e^{-e^x+2x} dx & =\left.\frac{\partial^2}{\partial a^2}(\Gamma(a))\right|_{a=2} \\ & =\left.\frac{\partial}{\partial a}(\Gamma(a) \psi(a))\right|_{a=2} \\ & =\Gamma(a) \psi^{\prime}(a)+\left.\Gamma(a) \psi^2(a)\right|_{a=2} \\ & =\psi^{\prime}(2)+\psi^2(2) \\&= \frac{\pi^2}{6}+\gamma^2-2\gamma \end{aligned} $$

Leibniz’s Rule came to my mind suddenly and brought me $$\int_{-\infty}^{+\infty} x^n e^{-e^x+a x} dx = \frac{\partial^n}{\partial a^n}(\Gamma(a)) =\frac{d^{n-1}}{d a^{n-1}}(\Gamma(a) \psi(a ))=\sum_{k=0}^{n-1}\binom{n-1}{k} \Gamma^{(k)}(a) \psi^{(n-1-k)}(a) ,$$ which is not helpful and definitely needs your help. Your comments and help are highly appreciated.

Lai
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1 Answers1

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You can express the $n$-th derivative as $$\frac{d^n}{da^n}\Gamma(a)=\Gamma(a)\operatorname Y_n\left(\psi(a),\psi^{(1)}(a),\dots,\psi^{(n-1)}(a)\right).$$ Where $\operatorname Y_n$ is the complete exponential Bell polynomial.

By Faà di Bruno’s formula, in expanded form, one may write $$\frac{d^n}{da^n}\Gamma(a)=\Gamma(a)\sum_{m_1+2m_2+\dots+nm_n=n}\frac{n!}{m_1!m_2!\dots m_n!}\prod_{j=1}^n\left(\frac{\psi^{(j-1)}(a)}{j!}\right)^{m_j}.$$

Antony Theo.
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