# Tricky integral and nasty Maple

1. Aug 15, 2006

### dextercioby

Let's say i feed Maple with this baby

$$\int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx$$

After some thinking he gives me

$$\int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{1}{12}\left( \ln 2\right) \pi ^{2}+\frac{1}{12}\pi ^{2}-\frac{1}{12}\pi ^{2}\gamma+\frac{1}{2}\zeta \left( 1,2\right)$$

I say OK, even though i don't understand who $\zeta \left( 1,2\right)$ is. Is it Hurwitz zeta function ? If so, then it should diverge... Anyway, the same software tells me that

$$\zeta \left( 1,2\right) = -.\,93754\,82543$$

So i say OK again, even though i'm not satisfied with what zeta stands for...

But that's not the issue. The issue is that i'm putting my head to work to do that integral and i get

$$\int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{\pi ^{2}}{12}\left( 1-\gamma \right) -\sum_{k=0}^{\infty }\left[ \left( -\right) ^{k}\frac{\ln \left( k+1\right) }{\left( k+1\right) ^{2}}\right]$$

So i'm thinking that the nasty sum should equal the 2 terms above, the one with the natural logarithm and the other with the dubious zeta function. But when i feed Maple with the sum, he doesn't return anything...:surprised

So what the heck? Dumb software, i figure... He woudn't give me not even an aproximate numerical value of the sum...

So the question is: any ideas on how to evaluate the sum...?

Daniel.

2. Aug 15, 2006

### George Jones

Staff Emeritus
I had no problem with getting Maple to evaluate your sum. The two expressions are equal.

Sorry if these questions are silly, but ...

Did you use evalf?

Did you use add (not sum)?

3. Aug 15, 2006

### dextercioby

I didn't use standard Maple,but an ancient version (5.3 i guess) implemented in Scientific WorkPlace 2.5.

I don't write code, i drag & drop symbols, so your questions, unfortunately don't help...

BTW, is that Hurwitz zeta function, or not...? If so, then it must be the notation from Andrews, Askey & Roy's book that would apply and not the one from Whittaker & Watson. The difference btw the 2 is the reversed order of arguments. However, let's say we adopt the former notation. Then

$$\zeta (1,2) =\zeta (2)= \frac{\pi^{2}}{6}\neq 0.9375...$$

So it's still unclear...

Daniel.

4. Aug 15, 2006

### benorin

Excerpt from Maple v10 help files:

The call Zeta(n, z) gives the nth derivative of the Zeta function, Zeta(n,z) = diff(Zeta(z), z\$n)

hence $$\zeta(1,2) = \zeta^{\prime}(2)$$

5. Aug 15, 2006

### dextercioby

Thanks for the input on the zeta function.

Daniel.

6. Aug 15, 2006

### George Jones

Staff Emeritus
I'm a big fan of Scierntific Workplace, and I even use it to write posts for these forums. I have an ancient (Maple) version of Scientific Workplace, but not as ancient as your version.

My version evalutes numerically your expression, but I couldn't just use the greek letter gamma for this. I copied and pasted into your expression the gamma that Scientific Workplace spat out when I asked it to do the original integral.

I am not very familiar with these special functions, or with standard notations for them.

7. Aug 15, 2006

### benorin

zeta prime sums

$$-\sum_{k=0}^{\infty }\left[ \left( -\right) ^{k}\frac{\ln \left( k+1\right) }{\left( k+1\right) ^{2}} = \sum_{k=1}^{\infty }\left[ \left( -\right) ^{k}\frac{\ln \left( k\right) }{k^{2}}$$

is related to the derivative of the zeta function:

Let $$\zeta (s)=\sum_{k=1}^{\infty}\frac{1}{k^s}$$ and $$\hat{\zeta} (s)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^s}$$ since

$$\sum_{k=1}^{\infty}\frac{1}{k^s}+\sum_{k=1}^{\infty}\frac{(-1)^k}{k^s} = 2^{1-s}\sum_{k=1}^{\infty}\frac{1}{k^s}$$​

so that

$$\zeta (s)+\hat{\zeta} (s) = 2^{1-s} \zeta (s) \Rightarrow \hat{\zeta} (s) =\left( 1- 2^{1-s}\right) \zeta (s)$$​

differentiating w.r.t. s, we have

$$\frac{d}{ds}\hat{\zeta} (s) = \frac{d}{ds}\left[ \left( 1- 2^{1-s}\right) \zeta (s) \right]$$

or

$$\frac{d}{ds}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^s}= 2^{1-s}\log (2) \zeta (s) + \left( 1- 2^{1-s}\right) \zeta^{\prime} (s)$$

in terms of sums

$$\sum_{k=1}^{\infty}\frac{(-1)^k\log (k)}{k^s} = 2^{1-s}\log (2) \sum_{k=1}^{\infty}\frac{1}{k^s} + \left( 1- 2^{1-s}\right) \sum_{k=1}^{\infty}\frac{\log (k)}{k^s}$$

to get your sum, set s=2.

8. Aug 15, 2006

### dextercioby

Funny, i've done the same calculations by myself. Thx for the effort, though. And $\hat{\zeta}$ is commonly denoted by $\eta$ and is called the eta function of Dirichlet.

Daniel.