An integral related to the harmonic number Definitions
$\Gamma(s)$ is a Gamma function
Defined as $\Gamma(s+1)=s!$
$H_{n,s}=\sum_{k=1}^{n}\frac{1}{n^s}$
Fractional arguments between 0 and 1 is defined by the integral,
$$\int_0^1\frac{1-x^a}{1-x}dx=H_a$$
Example: $H_{\frac{1}{2}}=2-2\ln(2)$
We proposed: proof that,
$$-\frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}\cdot \frac{1-e^{-bx}}{1-e^{ax}}dx=a^{-s}H_{\frac{a}{b},s} $$
Where $s >0$
 A: Hint:
$$\frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}\cdot e^{-\beta x}=\frac{1}{\beta^s}$$
$$1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}$$
A: We have
$$ I= \int_0 ^{\infty}  x^{s-1}\cdot \frac{1-e^{-bx}}{1-e^{ax}} dx $$
$$ I= -\int_0 ^{\infty}  x^{s-1}\cdot \frac{e^{-ax} -e^{-(b+a)x}}{1-e^{-ax}} dx $$
$$ I= -\int_0 ^{\infty}  x^{s-1}\cdot [e^{-ax} -e^{-(b+a)x}]\cdot \sum_{k=0} ^{\infty} e^{-akx} dx $$
$$ I= -\sum_{k=0} ^{\infty} \int_0 ^{\infty} e^{-akx} \cdot  x^{s-1}\cdot \left[{e^{-ax} -e^{-(b+a)x}}\right]  dx $$ 
$$ = -\sum_{k=0} ^{\infty} \int_0 ^{\infty}  x^{s-1} \cdot \left[e^{-(a+1)kx} - e^{-(b+a+ka)x} \right] dx $$
$$ = -\sum_{k=0} ^{\infty}\left[ \int_0 ^{\infty}  x^{s-1} \cdot \left[e^{-(a+1)kx}  \right]dx- \int_0 ^{\infty}  x^{s-1} \cdot \left[e^{-(b+a+ka)x}  \right] dx \right]$$
letting $u =(a+1)kx$ in the first and integral and $u=(b+a+ka)x$ yields:
$$ I= \sum_{k=0} ^{\infty} \left[\int_0 ^{\infty} u^{s-1}\cdot \frac{e^{-u}}{(ak+b+a)^{s}}du - \int_0 ^{\infty} u^{s-1}\cdot \frac{e^{-u}}{(ak+a)^{s}}du \right] $$
$$ =\sum_{k=1} ^{\infty} \left[\frac{Γ(s)}{(ak+b)^{s}} - \frac{Γ(s)}{(ak)^{s}}\right]$$ 
$$ = Γ(s)\cdot a^{-s}\cdot \sum_{k=1} ^{\infty} \left[ \frac{1}{(k+\frac{b}{a})^{s}} - \frac{1}{(k)^{s}} \right]$$
Now, it remains to show that this sum is equal to $-H_{\frac{a}{b},s}$.
$$H_{\frac{a}{b}} = \int_0 ^1  \frac {1-x^{\frac{a}{b}}} {1-x} dx $$ 
$$ = \int_0 ^1 \sum _{k=0} ^{\infty} x^k \cdot (1-x^{\frac{a}{b}})dx
 $$
$$ \sum _{k=1} ^{\infty} \left[ \frac{1}{k} -\frac{1}{k+\frac{a}{b}}  \right] $$
Now, by definition: $$ H_{\frac{a}{b},s} = \sum _{k=1} ^{\infty} \left[ \frac{1}{k^s} -\frac{1}{(k+\frac{a}{b})^s}  \right] $$
And the result follows.
