Evaluation of 1+(1/3^3)+(1/5^3)+

[AlbertEinstein]

Evaluation of 1+(1/3^3)+(1/5^3)+...

Hello guys ,I need help on the following which I encountered during reading the book "i:The imaginary tale".
The author writes that Euler evauated that
<br /> \frac{1}{1^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3}+... = \frac{\pi^2}{4}\ln2 + 2 \int_0^{\pi/2} x ln[sin(x)] dx<br />

He makes the reference that the work can be found in
1) Raymond Ayout-"Euler and the zeta function"-American Mathematical Monthly (dec 1974:pg 1067-86) and
2) Ronald Calinger "Leonhard Euler: The first st. Petersburg years(1727-1741)" Historica Mathematica 23 may 1996 pg 121-166

Since I have access to none of these I am unable to understand how he did it.I shall be grateful to you guys if you could explain or direct me to a webpage for explanation.

 

[shmoe]

I'm not positive how Euler actually did it, but it have been been similar to this:

First replace \sin{x} with

\frac{e^{ix}-e^{-ix}}{2i}=\frac{e^{ix}}{2i}(1-e^{-2ix})

Your integral becomes:

\int_0^{\pi/2}ix^2-x\log{2i}+x\log(1-e^{-2ix})dx

use the power series for log:

\log(1-y)=-y-y^2/2-y^3/3-\ldots

This is actually convergent on |y|=1, except when y=1.

Integrate this series term by term (integrate by parts) and do some rearranging. You'll get the sum you are after 1/1^3+1/3^3+... as well as

\frac{-1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}+\ldots=-\frac{\pi^2}{12}

which follows from 1+1/2^2+1/3^2...=pi^2/6, which Euler knew.

That should do it. I've seen a similar "proof" using the above steps to get this last result, namely 1+1/2^2+...=pi^2/6, but I can't remember if it was blamed on Euler or not There are some obvious holes that can be filled but are annoying, like justifying the term by term integration of this series where it's not absolutely convergent. Euler may have been able to make leaps like this without falling off cliffs, but most of us can't. I'll try to remember where I saw this before so you can see the details..

 

[AlbertEinstein]

The other way out..

Thanks Shmoe for your help. However the clarification you made was the other way out, i.e. you decomposed the integral into the infinite series.
But I don't think that Euler did it this way. I shall be very thankful if you could show how the integral came from the series. I hope you understand.
Thanks again.

 

[shmoe]

I found where I first saw a version of the above, near the end of:

http://www.american.edu/cas/mathstat/People/kalman/pdffiles/Sixways.pdf

and they cite Dennis Russell. The version given in the above it is much more obvious how you would go from sum to integral in the case they consider (dealing with the sum of 1/n^2). Given Russell's approach, it wouldn't have been difficult to proceed how I've outlined (though in reverse) to derive the integral.Alas, this doesn't appear to be how Euler did it, though he did have all the necessary tools I believe, or at least the gumption to assume the necessary results. I've looked up your Ayoub reference. It's pretty hefty and would take much typing to explain here. Is there no way you can look up the journal? AMM is a pretty common one, most colleges or universities probably carry it, or alternatively might have access to an electronic version via jstor. (I didn't look at Callinger)

Actually all of Ayoub is worth a read, not just this part. Euler's pre-Riemann work regarding the zeta function is always mentioned in passing, but this is the most detail I've looked at. A nice outline of some math history. An aside- it makes me wish I could read Latin and get something from Euler's work directly.

 

[AlbertEinstein]

Thanks.

Thanks again Shmoe. I have gone through that pdf. It was good and I understood most of them.Thanks for your help.