A shortcut to $\int_{0}^{\infty}\left ( 3e^{3x}\text{Ei}(-x)^4-4e^x\text{Ei}(-x)^3\text{Ei}(x) \right ) \text{d} x=-\frac{\pi^4}{10}$

Question

How can we prove this canonically? $$ \newcommand{\Ei}{\operatorname{Ei}} \newcommand{\Ci}{\operatorname{Ci}} \newcommand{\si}{\operatorname{si}} \newcommand{\Li}{\operatorname{Li}_4} \int_{0}^{\infty}\left ( 3e^{3x}\Ei(-x)^4-4e^x\Ei(-x)^3\Ei(x) \right ) \text{d} x=-\frac{\pi^4}{10}. $$ Here $\Ei$ is well-known to be exponent integral function. One proof is to follow the strategy and find them separately, $$ \begin{aligned} &\int_{0}^{\infty}e^{3x}\Ei(-x)^4\text{d}x = \frac{26\pi^4}{135} +\frac{4\pi^2}{9}\ln(2)^2 -\frac{4}{9}\ln(2)^4-\frac{32}{3}\Li\left ( \frac12 \right ) ,\\ &\int_{0}^{\infty}e^{x}\Ei(-x)^3\Ei(x)\text{d}x = \frac{61\pi^4}{360} +\frac{\pi^2}{3}\ln(2)^2 -\frac{1}{3}\ln(2)^4-8\Li\left ( \frac12 \right ), \end{aligned} $$ where $\operatorname{Li}_n(z)$ are polylogarithms. However, the simplicity of the result reminds me that a shortcut potentially exists. And I use this equality to find $$ \int_{0}^{\infty} \left ( \Ci(x)\cos(x)+\si(x)\sin(x) \right ) \left ( \si(x)\cos(x)-\Ci(x)\sin(x)\right )^4\text{d}x= -\frac{\pi^5}{160}, $$ where $\Ci\left ( x \right ) =-\int_{x}^{\infty} \frac{\cos t}{t} \text{d}t$ (resp. $\si\left ( x \right ) =-\int_{x}^{\infty} \frac{\sin t}{t} \text{d}t$) is cosine(resp. sine) integral function. This is intriguing too. And I'm thankful to your suggestions and comments.

Answer 1

In my notations here, the trigonometric integral could be done by symmetry. $$ \int_{0}^{\infty} A(x)B(x)^4\text{d}x =-\int_{[0,\infty]^5} \frac{1}{\left ( 1+x_1^2\right ) \left ( 1+x_2^2\right )\left ( 1+x_3^2\right ) \left ( 1+x_4^2\right )\left ( 1+x_5^2\right ) } \frac{x_1}{x_1+x_2+x_3+x_4+x_5}\text{d}x_1...\text{d}x_5$$ $$ =-\frac{1}{5} \left ( \frac{\pi}{2} \right )^5. $$ And by contour integral reformulation proves our claim. A generalization for the $7$-dimensional integral is given by $$ \newcommand{\Ei}{\operatorname{Ei}} \int_{0}^{\infty} \left (5e^{5x}\Ei(-x)^6 -18e^{3x}\Ei(-x)^5\Ei(x) +5e^x\Ei(-x)^4\left ( 3\Ei(x)^2- \pi^2\right ) \right ) \text{d}x =\frac{\pi^6}{14}, $$ which shows certain $\mathbb{Q}$-linear dependency among these exponential periods.