one pound

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@vv thank you for the detailed answer I'm sure this will prove useful in my learning maple in general.

@dharr I appear to have copied the wrong pic from wiki. It is indead the infinite version I require - thank you for pointing that out; I've updated the question with the correct function.

thanks guys will let you know how I get on with more difficult  sums.

@Mariusz Iwaniuk thanks that works well but according to https://en.wikipedia.org/wiki/Exponential_integral it is not accurate for x>2.5. I need to use it in an asymptotic estimation. 

wiki suggests {\displaystyle E_{1}(z)={\frac {\exp(-z)}{z}}\sum _{n=0}^{N-1}{\frac {n!}{(-z)^{n}}}} but according to vv 3894 its no good as it diverges but I wonder does not the asymptotic expansion of Ii(x) do the same but is still useful at a single point.

@vv @vv 3889  would it be okay or worst for asymptotic estimations?

@Mariusz Iwaniuk yes thank you for looking into this further. I have been looking at mathematica too with its progammable element but I'm not sure it would help in the end with

int ln(a-x)/ln(a+x) dx

expressed as a power series in sigma notation so I think I will just have to work with the summands for now. Might look into mathematica on mathematica.stackexchange.com

@Mariusz Iwaniuk okay that makes sense regarding the integral but I'm not sure why it at least does not do:

convert(ln(a-x)/ln(a+x), FormalPowerSeries, x)

to give a sigma for this. 

@Mariusz Iwaniuk 264 it is not quite working for

convert(int(ln(a-x)/ln(a+x), x), FormalPowerSeries, x)

for some reason yet the series does expand properly

series(int(ln(a-x)/ln(a+x), x), x = 0, 8) 

@Mariusz Iwaniuk 264  Thank you for the suggestions. When it comes to expressions involving

ln(x) 

 

is there anyway to force the series

series(int(ln(a+x)/a, x), x = 0, 8)

to do the same with does commands? turn it into sigma notation?

@vv 3212 thanks, yes I've run that (see below) and it seems to work. But why does setting t = exp(1) ..  x make the difference?

with(IntegrationTools);

n := 5; q := Int(1/ln(t), t = exp(1) .. x);

for k to n do

q := simplify(Parts(q, GetIntegrand(q)))

end do;

@rlopez & @rlopez 2081 rlopez 2081 perhaps the part can be made to alternate giving the integration by parts expansion?

@Axel Vogt I see, yes thank you for making that clear. I was hoping for an integration by parts asymptotic expansion of Li(x) amonst othe functions.

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