In my quest to solve this complicated integral and learning about the Meijer function, and its variations, I am trying to compute the following:
$$\int_{0}^{\infty} x^{\alpha}e^{-\beta x}\ln(1+x)\mathrm dx$$
$\alpha,\beta > 0$ are real values.
I have tried to use integration by part but it lead me to nowhere (incomplete Gamma function divided by (1+x). I also tried to transform using the Meijer function as $\ln(1+x) =G_{2,2}^{1,2}\left(\left.\begin{array}{l|l}1,1 \\ 1,0\end{array} \right\rvert\, x\right)$ but then I got stuck in how to continue the integral. I looked into previous posts but they are incomplete. I reached books such as integrals and series with no success.
Can someone enlighten me please on how to solve this obscure thing?
PS: Is there a way to solve this in case it becomes $\int_{0}^{\infty} x^{\alpha}e^{-\beta x^c}\ln(1+x)\mathrm dx$ ??
Best Answer
Denote $$I(\alpha, \beta) := \int_0^\infty x^\alpha e^{-\beta x} \log(1 + x) \,dx.$$
A (relatively) nice expression for $I(\alpha, \beta)$ is available for the case that $\alpha$ is a nonnegative integer: Integrating by parts and then substituting $u = 1 + x$ gives $$I(0, \beta) = \int_0^\infty e^{-\beta x} \log(1 + x) \,dx = \frac1\beta \int_0^\infty \frac{e^{-\beta x} \,dx}{1 + x} = \frac{e^\beta}{\beta} \int_1^\infty \frac{e^{-\beta u} \,du}{u} = \frac{e^\beta}{\beta} \operatorname{E}_1(\beta) ,$$ where $\operatorname{E}_1(x) := \int_x^\infty \frac{e^{-t}}{t} \,dt$ is the exponential integral function. Now, differentiating both sides $\alpha$ times with respect to $\beta$ gives $$ I(\alpha, \beta) = (-1)^\alpha \frac{d^\alpha}{d\beta^\alpha} I(0, \beta) = (-1)^\alpha \frac{d^\alpha}{d\beta^\alpha} \left(\frac{e^\beta}{\beta} \operatorname{E}_1(\beta)\right) . $$ Direct application of the general Leibniz rule then gives the explicit formula $$I(\alpha, \beta) = \left[\sum_{k = 0}^\alpha \frac{(-1)^{\alpha + k + 1} {}_\alpha P_k}{\beta^{k + 1}} \right] e^\beta \operatorname{E}_1(\beta) + \sum_{j = 1}^\alpha \sum_{k = 0}^{\alpha - j} \sum_{l = 0}^{j - 1} \frac{(-1)^{\alpha + j + k} {}_{\alpha - j} P_k \cdot {}_{j - 1} P_l}{\beta^{j + k + 2}},$$ where as usual ${}_n P_k = \frac{n!}{(n - k)!}$.
For example, \begin{align*} I(1, \beta) &= \left(\frac1{\beta^2} - \frac1\beta\right) e^\beta \operatorname{E}_1(\beta) + \frac{1}{\beta^2} \\ I(2, \beta) &= \left(\frac2{\beta^3} - \frac2{\beta^2} + \frac1\beta\right) e^\beta \operatorname{E}_1(\beta) + \left(\frac3{\beta^3} - \frac1{\beta^2}\right) . \end{align*}
Somewhat interestingly, if we fix $\beta = 1$, so that $$I(\alpha, 1) = p_\alpha e \operatorname{E}_1(1) + q_\alpha,$$ for integers $p_\alpha, q_\alpha$, then
These facts suggest there might be a combinatorial argument for evaluating $I(\alpha, 1)$.
I don't know offhand how to produce it, but Maple gives an unwieldy formula for general $\alpha$: \begin{multline*}I(\alpha, \beta) \\ = \frac1{\beta^{1 + \alpha}} \Bigg(\left[ (-1)^{-1-\alpha} \pi \alpha \Gamma (\alpha, -\beta) + (-1)^{-\alpha} \pi \Gamma (1 + \alpha) - \frac{\pi e^\beta}{\beta}\right] \csc \pi \alpha \\ \qquad \qquad \qquad - \pi \Gamma (1 + \alpha) \cot \pi \alpha + \Psi (-\alpha) \Gamma (1 + \alpha) - \Gamma (1 + \alpha) \log \beta \Bigg) \\ + \frac1{\beta^\alpha} {}_2 F_2 (1, 1; 2, 1 - \alpha; \beta) \Gamma(\alpha) .\end{multline*} Here $\Gamma(\,\cdot\,,\,\cdot\,)$ is the incomplete Gamma function, $\Psi$ is the Digamma function, and ${}_2 F_2$ is a (generalized) hypergeometric function.