Take the 2-minute tour ×
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.

How to compute $\int\frac{x^7}{\sin(x)} dx$ efficiently ? We need $Polylog$ for this.

share|improve this question
    
Not sure I understand. Do you need to efficiently evaluate the definite version of the integral? Where is efficiency in the analytical solution? –  gt6989b Sep 21 '12 at 19:42
    
Efficients as in not too complicated or long. Efficient as in not 7 substitutions and 10 partials used. –  mick Sep 21 '12 at 19:46
3  
The most efficient way today is to ask wolfram –  no identity Sep 21 '12 at 19:54
1  
Do you have any reason to think there is a less nasty answer? Maple, by the way, gives what appears to be the same answer. –  Robert Israel Sep 21 '12 at 20:05
2  
OK then, the short method is to ask Wolfram or Maple. –  Robert Israel Sep 21 '12 at 20:09

1 Answer 1

up vote 5 down vote accepted

You could substitute $w = e^{ix}$, obtaining $2 i \int \log(w)^7/(w^2-1)\ dw$. Now $$ 2 i \int \frac{w^p\ dw}{w^2-1} = -i{w}^{p+1}{\Phi} \left( {w}^{2},1,\frac{p+1}{2} \right)$$ (where $\Phi$ is the Lerch Phi function) so take the $7$'th derivative of this with respect to $p$ and evaluate at $p=0$. Then substitute back $w = e^{ix}$. According to Maple the result is $$ -\frac{7}{2}\,{{\rm e}^{ix}} \left( \frac{2}{7}\,{x}^{7}{\Phi} \left( {{\rm e} ^{2\,ix}},1,1/2 \right) +i{x}^{6}{\Phi} \left( {{\rm e}^{2\,ix }},2,1/2 \right) -3\,{x}^{5}{\Phi} \left( {{\rm e}^{2\,ix}},3, 1/2 \right) \right.\\ -\frac{15}{2}\,i{x}^{4}{\Phi} \left( {{\rm e}^{2\,ix}},4,1 /2 \right) +15\,{x}^{3}{\Phi} \left( {{\rm e}^{2\,ix}},5,1/2 \right) +{\frac {45}{2}}\,i{x}^{2}{\Phi} \left( {{\rm e}^{2\, ix}},6,1/2 \right)\\\left. -{\frac {45}{2}}\,x{\Phi} \left( {{\rm e}^{ 2\,ix}},7,1/2 \right) -{\frac {45}{4}}\,i{\Phi} \left( { {\rm e}^{2\,ix}},8,1/2 \right) \right) $$

share|improve this answer

Your Answer

 
discard

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.