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.
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