MHB An abomination of a series.... Can anyone help?

[DreamWeaver]

Hi all! (Sun)

I'm utterly at a loss here, so I was wondering if one of you kind souls would, or even could, for that matter, take pity on me and help out on this exceedingly vexatious problem...?Define the Second Order Clausen function by:$$\text{Cl}_2(x) = \sum_{k=1}^{\infty} \frac{\sin kx}{k^2}$$And the Laplace transform of a function $$f(x)$$ by:$$\mathfrak{L}(f) = \int_0^{\infty} e^{-wx}f(x)\, dx = F(w)$$I've been trying to evaluate the Laplace Transform of the Clausen function. Here's what I've got so far:$$\mathfrak{L}(\text{Cl}_2(x)) = \int_0^{\infty} e^{-wx}\, \text{Cl}_2(x)\, dx = $$$$\sum_{k=1}^{\infty} \frac{1}{k^2}\, \int_0^{\infty} e^{-wx}\sin kx\, dx $$As I've shown in the Laplace Transform tutorial thread (see post #3 here http://mathhelpboards.com/math-notes-49/laplace-transforms-proofs-10892.html ):For $$\mathscr{Re}(w) > |\mathscr{Im}(a)|$$$$\mathfrak{L}(\sin ax) = \frac{a}{w^2+a^2} $$[NB. I forgot to add these conditions in the tutorial, and now it's too late to edit. ]Since the summation index, $$k$$, of the Clausen function, $$\text{Cl}_2(x)$$, is a nonzero, positive integer $$k\in\mathbb{N}$$, then $$\mathfrak{L}(\text{Cl}_2(x)) = \sum_{k=1}^{\infty} \frac{1}{k^2}\, \left[ \frac{k}{k^2+w^2}\right] = \sum_{k=1}^{\infty} \frac{1}{k(k^2+w^2)}$$The thing is, I'm sure I've seen series like this before, but for the life of me, I can't think where, or in what context... Can anyone help?Incidentally, at the very least, my efforts so far aren't entirely fruitless, as I have the inverse Laplace Transform$$\mathfrak{L}\left( \sum_{k=1}^{\infty} \frac{1}{k(k^2+w^2)} \right) = \text{Cl}_2(x)$$It's pretty slim comfort, mind... (Worried)(Doh)

 

[DreamWeaver]

Alternatively, I've also been considering the Clausen function of First Order:$$\text{Cl}_1(x) = -\log \Bigg| 2\sin \frac{x}{2} \Bigg| = \frac{d}{dx}\, \text{Cl}_2(x) = \sum_{k=1}^{\infty}\frac{\cos kx}{k}$$Here's what I get:$$\mathfrak{L}(\text{Cl}_1(x)) = \sum_{k=1}^{\infty}\frac{1}{k}\, \int_0^{\infty}e^{-wx}\cos kx\, dx = \sum_{k=1}^{\infty} \frac{1}{k} \left[ \frac{w}{k^2+w^2} \right]$$Again, I gen an abominable series that looks annoyingly familiar. Grrrr... (Headbang)(Headbang)(Headbang)

 

[alyafey22]

We can write the summand as

$$\frac{1}{k(k^2+w^2)}=\frac{i}{2w}\left( \frac{1}{k(k+iw)}-\frac{1}{k(k-iw)}\right)$$

Then use the digamma function.

 

[DreamWeaver]

We can write the summand as

$$\frac{1}{k(k^2+w^2)}=\frac{i}{2w}\left( \frac{1}{k(k+iw)}-\frac{1}{k(k-iw)}\right)$$

Then use the digamma function.

Ha ha! What a silly mammal! [Me, that is, not you] Thanks mate! I'll give that a whirl... (Handshake)

Would be nice to nail the Laplace Transform of the Clausen function. Especially for a Clausen-fetishist like myself... (Bandit)

Cheers! (Sun)

 

[alyafey22]

You can then use the recurrence and reflection formulas to obtain

$$\psi(1+x)-\psi(1-x)=\frac{1}{x}-\pi \cot (\pi x)$$

 

[DreamWeaver]

Also, on the subject of strange, unusual, insane, and mildly titillating Laplace Transforms, this one also looks quite interesting/scary/arousing/abominable [delete as applicable - though I'll delete none of them! None of them, I tell ya! Arrr... ].Define the Dilogarithm in the usual way (for $$-1 \le x \le 1$$):$$\text{Li}_2(x) = -\int_0^1\frac{\log(1-xt)}{t}\, dt = \sum_{k=1}^{\infty}\frac{x^k}{k^2}$$Then$$\mathfrak{L}(\text{Li}_2(x)) = \sum_{k=1}^{\infty}\frac{1}{k^2}\, \int_0^{\infty} x^k e^{-wx}\, dx = $$$$\sum_{k=1}^{\infty}\frac{1}{k^2}\, \left[ \frac{1}{w}\, \int_0^{\infty}e^{-t}\left( \frac{t}{w} \right)^k \, dt \right] = $$
$$\sum_{k=1}^{\infty}\frac{(1/w)^{k+1}\, \Gamma(k+1) }{k^2} = \sum_{k=1}^{\infty}\frac{k!\, (1/w)^{k+1} }{k^2} = \sum_{k=1}^{\infty}\frac{(k-1)!\, (1/w)^{k+1} }{k}$$
$$\therefore\quad \mathfrak{L}(\text{Li}_2(x)) = \sum_{k=1}^{\infty}\frac{(k-1)!\, (1/w)^{k+1} }{k}$$
Might be worth a look... (Bandit)

 

[DreamWeaver]

You can then use the recurrence and reflection formulas to obtain

$$\psi(1+x)-\psi(1-x)=\frac{1}{x}-\pi \cot (\pi x)$$

Thanks, Zaid! That's another one I hadn't thought of... I really appreciate all your help! (Handshake)(Hug)

 

[alyafey22]

The Laplace transform of the dilog seems interesting. I think the resultant series could be written in terms of the hyper geometric function. I have to go to sleep now , will look at later :)

 

[DreamWeaver]

I have to go to sleep now , will look at later :)

Totally understood. Sweet dreams. (Sleepy) :D

The Laplace transform of the dilog seems interesting. I think the resultant series could be written in terms of the hyper geometric function.

That's a very good idea... Hmmm! (Thinking)

 

[alyafey22]

I must have overlooked that but I think it is not possible to integrate the series representation since that is only valid in the unit circle. We might have luck for the integral representation but we have to prove that it converges and we can extend it behind $x>1$. I think that is possible since Li(2) is well-defined but it will give complex values.

 

[alyafey22]

The series in the transform of the Clausen function could also be solved using complex analysis.

 

[DreamWeaver]

I must have overlooked that but I think it is not possible to integrate the series representation since that is only valid in the unit circle. We might have luck for the integral representation but we have to prove that it converges and we can extend it behind $x>1$. I think that is possible since Li(2) is well-defined but it will give complex values.

Yes! In the cold light of day, it looks far less appetizing... (bh)(bh)(bh)

 

[alyafey22]

Hey Dw , have you tried the Laplace transform of

$\text{Li}_2(e^{-t})$ and $\text{Li}^2_2(e^{-t})$

I think they are doeable. And possibly the generalization $\text{Li}^n_p(e^{-t})$.