Mellin trasform $\text{csch}^3(x)$ Could you show the following integral ?:
$$
\int_{0}^{\infty}x^{s - 1}\operatorname{csch}^{3}\left(x\right)\,\mathrm{d}x =
2^{-s}\left[\left(2^{s} - 4\right)\zeta\left(s - 2\right) -
\left(2^{s} - 1\right)\zeta\left(s\right)\right]\Gamma\left(s\right)$$
 A: So here is my attempt, I don't know what it will be but in any case it will be, in my opinion, a great starting point.
First of all
$$\text{cosech}(x) = \frac{2}{e^x - e^{-x}}$$
Hence
$$\int_0^{+\infty} x^{s-1}\ \left(\frac{2}{e^x - e^{-x}}\right)^3\ \text{d}x$$
Arranging a bit to get
$$8\int_0^{+\infty}x^{s-1} \frac{(e^x)^3}{(e^{2z}-1)^3}\ \text{d}x =8\int_0^{+\infty}x^{s-1} \frac{e^{3x}}{(e^{2x}(1 - e^{-2x})^3} = 8\int_0^{+\infty}x^{s-1} \frac{e^{3x}}{e^{6x}(1 - e^{-2x})^3}\ \text{d}x$$
Now we can use the Geometric Series:
$$\frac{1}{(1 - e^{-2z})^3} = \frac{1}{2} \sum_{k = 2}^{+\infty} k(k-1)(e^{-2x})^{k-2}$$
to get
$$8\int_0^{+\infty}x^{s-1} e^{-3x} \frac{1}{2} \sum_{k = 2}^{+\infty} k(k-1)(e^{-2x})^{k-2} = 4\sum_{k = 2}^{+\infty} k(k-1)\int_0^{+\infty}x^{s-1} e^{-3x} (e^{-2x})^{k-2}\ \text{d}x $$
So
$$4\sum_{k = 2}^{+\infty} k(k-1)\int_0^{+\infty}x^{s-1} e^{-3x} e^{-2x(k-2)}\ \text{d}x = 4\sum_{k = 0}^{+\infty} k(k-1)\int_0^{+\infty}x^{s-1} e^{-x(2k-1)}\ \text{d}x$$
Now this is a well known integral: it's indeed:
$$\int_0^{+\infty}x^{s-1} e^{-x(2k-1)}\ \text{d}x = (2k-1)^{-s}\Gamma(s)$$
Indeed the two necessary conditions are satisfied:
$$\Re(s) >0 ~~~~~~~ \Re(k) > \frac{1}{2}$$
Finally we end up with a series:
$$4\sum_{k = 2}^{+\infty} k(k-1)(2k-1)^{-s}\Gamma(s) = 4\Gamma(s) \sum_{k = 2}^{+\infty} k(k-1)(2k-1)^{-s}$$
The series
$$\boxed{\sum_{k = 2}^{+\infty} k(k-1)(2k-1)^{-s}}$$
is where I do stop for the moment, because I need to check if what I did makes sense, and I need to find a sum for that...
Possible details, later
FINAL EDIT, THANKS TO USER1952009 (you can see his comment below!)
The series can start from $k = 1$, since the very first term will be zero. 
Then, using a cute trick: write the part
$$4k(k-1) = 4k^2 - 4k$$
as
$$(2k-1)^2 - 1$$
To get
$$\sum_{k = 1}^{+\infty}\left((2k-1)^2 - 1\right)(2k-1)^{-s}$$
Splitting the sum into two pieces, we recognise the Riemann Zeta:
$$\sum_{k = 1}^{+\infty} (2k-1)^{2-s} = 2^{2-s} \left(2^{s-2}-1\right) \zeta (s-2)$$
$$\sum_{k = 1}^{+\infty} (2k-1)^{-s} = 2^{-s} \left(2^s-1\right) \zeta (s)$$
Putting together:
$$2^{2-s} \left(2^{s-2}-1\right) \zeta (s-2) - 2^{-s} \left(2^s-1\right) \zeta (s) = $$
This can be transformed, with simple algebra into
$$2^{-s}\left((2^2 - 4)\zeta(s-2) - (2^s-1)\zeta(s)\right)$$
Without forgetting the Gamma term we left apart, we end up with your desired result:
$$\boxed{2^{-s}\left((2^2 - 4)\zeta(s-2) - (2^s-1)\zeta(s)\right)\Gamma(s)}$$
