including non-elementary functions

 
It is well known that the following functions cannot be expressed in terms of a finite number of elementary function.
$\int_0^{\infty} e^{-t}t^{x-1}dt,~~~~\int_{0}^{x}{e^{-t^2}}dt,~~~~~\int_{-x}^{\infty}\frac{e^{-t}}{t}dt,~~~~~\int_{0}^{x}\frac{\sin{t}}{t}dt,~~~~~\int_{x}^{\infty}\frac{\cos{t}}{t}dt,~~~~~\int_{0}^{x}\frac{1}{\log{t}}dt$
 
Then they give rise to the functions
$\Gamma(x)\equiv\int_0^{\infty} e^{-t}t^{x-1}dt$ (x>0)
${\rm erf}(x)\equiv\frac{2}{\sqrt{\pi}}\int_{0}^{x}{e^{-t^2}}dt$
${\rm Ei}(x)\equiv-\int_{-x}^{\infty}\frac{e^{-t}}{t}dt$
${\rm Si}(x)\equiv\int_{0}^{x}\frac{\sin{t}}{t}dt$
${\rm Ci}(x) \equiv -\int_{x}^{\infty}\frac{\cos{t}}{t}dt$
${\rm li}(x)\equiv\int_{0}^{x}\frac{1}{\log{t}}dt$
 
C is a constant of integration.
 
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$\log{\Gamma(x)}=(x-\frac{1}{2})\log{x}-x+\frac{\log{2\pi}}{2}+2\int_{0}^{\infty} \frac{\arctan{(\frac{t}{x})}}{e^{2\pi t}-1} dt$
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$\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}=\int_0^1 x^{p-1}(1-x)^{q-1}dx$
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$\Gamma(1+\frac{x}{2})\Gamma(\frac{1}{2}-\frac{x}{2})({\prod_{n=1}^{\infty}(1-\frac{x}{2n-1})(1+\frac{x}{2n})})=\sqrt{\pi}$
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$\Gamma(2x)=\frac{2^{2x-1}}{\sqrt{\pi}}\Gamma(x)\Gamma(x+\frac{1}{2})$
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$\int_{0}^{1}\log{\Gamma(x+1)}dx=\frac{1}{2}\log{(2\pi)}-1$
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$\int_{0}^{\infty} \frac{dx}{\Gamma(a+x)\Gamma(a-x)}=\frac{2^{2a-3}}{\Gamma(2a-1)}~~~~~~~~~~(a>\frac{1}{2})$
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$\Gamma(x)=\lim_{n \rightarrow \infty} \frac{n!n^x}{\displaystyle \prod_{i=0}^{n}(x+i)}$
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$\Gamma(x) = \frac{1}{x} \prod_{n=1}^{\infty} (1+\frac{1}{n})^x(1+\frac{x}{n})^{-1}$
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$(2\pi)^{\frac{n-1}{2}}\Gamma(x)=n^{x-\frac{1}{2}}\prod_{i=1}^{n}\Gamma(\frac{x+i-1}{n})$
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$\frac{1}{\Gamma(z)}=ze^{\displaystyle z\lim_{n\rightarrow\infty}(\sum_{k=1}^{n}\frac{1}{k}-\log{n})}\prod_{m=1}^{\infty}(1+\frac{z}{m})e^{-\frac{z}{m}}$
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$\int_{0}^{1}x^p(\log\frac{1}{x})^qdx=\frac{\Gamma(q+1)}{(p+1)^{q+1}}~~~~~~~~~~(p,q>-1)$
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$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin{({\pi}x)}}~~~~~~~~~~(0
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$\int_{0}^{1} \log{\Gamma(x)}dx =\frac{1}{2}\log{(2\pi)}$
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$\Gamma(\frac{2}{3})\Gamma(\frac{1}{3})=\frac{2\pi}{\sqrt{3}}$
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$\Gamma(\frac{3}{4})\Gamma(\frac{1}{4})=\pi\sqrt{2}$
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$\Gamma(\frac{4}{5})\Gamma(\frac{1}{5})=\frac{\pi\sqrt{2}\sqrt{5+\sqrt{5}}}{\sqrt{5}}$
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$\Gamma(\frac{7}{8})\Gamma(\frac{1}{8})=\pi2^{\frac{3}{4}}\sqrt{\sqrt{2}+1}$
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$\int{\rm erf}(x)dx=\frac{e^{-x^2}}{\sqrt{\pi}}+x{\rm erf}(x)+C$
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\int{x{\rm erf}(x)}dx=\frac{x^2}{2}{\rm erf}(x)-\frac{1}{4}{\rm erf}(x)+\frac{xe^{-x^2}}{2\sqrt{\pi}}+C
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$\int{\rm Si}(x)dx=x{\rm Si}(x)+\cos{x}+C$
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$\int{\rm Ci}(x)dx=x{\rm Ci}(x)-\sin{x}+C$
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$\lim_{x\rightarrow\infty}{\rm Si}(x)=\frac{\pi}{2}$
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$\lim_{x\rightarrow\infty}{\rm Ci}(x)=0$
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$\int{\rm li}(x)dx = x{\rm li}(x)-{\rm Ei}(2\log{x})+C$
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$\int_{0}^{1}{\rm li}(x)dx = -\log{2}$

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$\int_{0}^{1}{\rm li}(\frac{1}{x})xdx = 0$

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$\int_{0}^{p}{\rm Ei}(ax)dx = p {\rm Ei}(ap)+\frac{1-e^{ap}}{a}$

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${\rm li}(x)={\rm Ei}(\log{x})$