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I think these should hav some closed form: $$\displaystyle\begin{align*} & \int_{0}^{1}{\frac{\left( 1-x \right)\ln \left( x \right){{\text{e}}^{-x}}}{\pi -x}}\text{d}x \\ & \int_{0}^{\infty }{\frac{1}{x{{\text{e}}^{x}}\left( {{\pi }^{2}}+{{\ln }^{2}}x \right)}}\text{d}x \\ \end{align*}$$

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I am not optimistic about that. – André Nicolas Jan 28 '13 at 8:20
That's a statement. Not a question. – mrf Jan 28 '13 at 8:24
Where did these integrals come from? – Mhenni Benghorbal Jan 28 '13 at 9:16
@MhenniBenghorbal : These are collected by my friend – gauss115 Jan 28 '13 at 9:48
up vote 1 down vote accepted

I don't know what is special about $\pi$ in your integrals. In the first, if we replace $\pi$ by $t > 1$, we get

$$\eqalign{J(t) &= \int_0^1 \frac{(1-x) \ln(x) e^{-x}}{t - x} \ dx = \sum_{n=0}^\infty t^{-n-1} \int_0^1 (1-x) \ln(x) e^{-x} x^n\ dx \cr&= \sum_{n=0}^\infty t^{-n-1} (G(n) - G(n+1))\cr &= t^{-1} G(0) + (1/t - 1) \sum_{n=1}^\infty t^{-n} G(n)}$$ where $$ \eqalign{G(s) &= \int_0^1 \ln(x) e^{-x} x^s \ dx = \dfrac{d}{ds} \int_0^1 e^{-x} x^s \ dx\cr &= \dfrac{d}{ds} {\frac {{\mbox{$_1$F$_1$}(1;\,2+s;\,1)}{{\rm e}^{-1}}}{s+1}}\cr &= -{\frac {{\mbox{$_1$F$_1$}(1;\,2+s;\,1)}{{\rm e}^{-1}}}{ \left( s+1 \right) ^{2}}}-{\frac { {\mbox{$_2$F$_2$}(2+s,2+s;\,3+s,3+s;\,-1)}}{ \left( s+1 \right) \left( 2+s \right) ^{2}}} }$$ But I don't know how to get a closed form for the sum.

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