Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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

1 Answer 1

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.