ASKSAGE: Sage Q&A Forum - Individual question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 08 Aug 2015 04:06:06 -0500Simplify result of this definite integralhttp://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/It is well known that for $n\in\mathbb{N}$ and $n>0$ (an maybe even for more than these restrictions): $$I_n = \int_0^\infty\frac{x^n}{e^x-1}dx = \zeta(n+1)n!$$ which, analytically can be shown easily by expanding $1/(1-e^{-x})$ into a geometric series, which leads to trivial integrals, and by using $\zeta(n+1)=\sum_{l=1}^\infty l^{-(n+1)}$. So, eg.: $$I_1 = \pi^2/6$$ $$I_2 = 2\zeta(3)$$ a.s.o...
Now, if I try even the simplest case with sage, I get this 'nifty' little results
sage: integrate(x/(exp(x)-1),x,0,oo)
-1/6*pi^2 + limit(-1/2*x^2 + x*log(-e^x + 1) + polylog(2, e^x), x,
+Infinity, minus)
**Is there any trick to simplify this** down to the final result, or is this about as far as I can get with sage alone?
PS.: it is probably needless to say that (once again ... :( ...)
In[1]:= Integrate[x/(Exp[x] - 1), {x, 0, Infinity}]
Out[1]:= Pi^2/6
Wed, 04 Sep 2013 10:06:49 -0500http://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/Answer by Mark for <p>It is well known that for $n\in\mathbb{N}$ and $n>0$ (an maybe even for more than these restrictions): $$I_n = \int_0^\infty\frac{x^n}{e^x-1}dx = \zeta(n+1)n!$$ which, analytically can be shown easily by expanding $1/(1-e^{-x})$ into a geometric series, which leads to trivial integrals, and by using $\zeta(n+1)=\sum_{l=1}^\infty l^{-(n+1)}$. So, eg.: $$I_1 = \pi^2/6$$ $$I_2 = 2\zeta(3)$$ a.s.o...</p>
<p>Now, if I try even the simplest case with sage, I get this 'nifty' little results</p>
<pre><code>sage: integrate(x/(exp(x)-1),x,0,oo)
-1/6*pi^2 + limit(-1/2*x^2 + x*log(-e^x + 1) + polylog(2, e^x), x,
+Infinity, minus)
</code></pre>
<p><strong>Is there any trick to simplify this</strong> down to the final result, or is this about as far as I can get with sage alone?</p>
<p>PS.: it is probably needless to say that (once again ... :( ...)</p>
<pre><code>In[1]:= Integrate[x/(Exp[x] - 1), {x, 0, Infinity}]
Out[1]:= Pi^2/6
</code></pre>
http://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/?answer=28777#post-id-28777for an answer to this and related other questions see http://thingwy.blogspot.de/Sat, 08 Aug 2015 03:21:43 -0500http://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/?answer=28777#post-id-28777Comment by rws for <p>for an answer to this and related other questions see <a href="http://thingwy.blogspot.de/">http://thingwy.blogspot.de/</a></p>
http://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/?comment=28778#post-id-28778That's not an anwer fitting in this forum.Sat, 08 Aug 2015 04:06:06 -0500http://ask.sagemath.org/question/10503/simplify-result-of-this-definite-integral/?comment=28778#post-id-28778