Is there any closed form solution available for following integral?

Question

I want to solve following integral $$\int_{0}^{x}\int_0^{x}\cdots \int_0^{x}(Mx-x_1-x_2-\cdots x_M)\ln (Mx-x_1-x_2-\cdots x_M)e^{-x_1}\cdots e^{-x_M}dx_1\cdots dx_M$$ where $x$ and $M$ are positive constants. Any approximation or hint is also welcome. Thanks in advance.

Answer 1

Do the change of variables $x \to y$ with $y_1 = \sum_i x -x_i$, and $y_j = x-x_j \quad \forall j \ne 1$. The Jacobian of this change of variable is $1$, so the integral $\text{R}$ is equal to

$$ \text{R} = \frac{1}{\exp(Mx)} \int_{x}^{0} \dots \int_{x}^0 \int_{x + \sum_{i=2}^n y_i}^{\sum_{i=2}^n y_i} y_1 \ln y_1 \exp (y_1) dy_1 dy_2 \dots dy_{M} $$ Then you integrate by parts (derive $y_1ln(y_1)$, integrate $exp(y_1)$), and you can have an expression with the exponential integral (special function) since you need to integrate $ln(y_1)exp(y_1)$ (integrating by parts gives you $exp(y_1)/y_1$).

Since you then need to integrate the exponential integral (special function) afterwards, there is no closed form for this integral in my opinion.

OLD ANSWER: NOT RELEVANT, only gives a upper bound on the integral. Do the change of variables $x \to y$ with $y_1 = \sum_i x -x_i$, and $y_j = x-x_j \quad \forall j \ne 1$. The Jacobian of this change of variable is $1$, so the integral $\text{R}$ is smaller than (the integration domain is to big, compared to the right integral) $$ \text{R} \le \frac{1}{\exp(Mx)} \int_{x}^{0} \dots \int_{x}^0 dy_2 \dots dy_{M} \int_{ Mx}^{0} y_1 \ln y_1 \exp (y_1) dy_1 $$ $$ = - \frac{(-x)^{M-1}}{\exp(Mx)} \int_{0}^{Mx} y_1 \ln y_1 \exp (y_1) dy_1 $$ Then you integrate by parts (derive $y_1ln(y_1)$, integrate $exp(y_1)$), and you can have an expression with the exponential integral (special function) since you need to integrate $ln(y_1)exp(y_1)$ (integrating by parts gives you $\exp(y)/y$).