<s>[INST] Can anyone show that $\int_0^1\left[\frac{B-1}{1-x}+\frac{B(x^B-x)}{(1-x)(1-x^B)}\right]\sum _{n=1}^{\infty}x^{B^n-1}dx=\gamma$ Euler's constant
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{k}-\ln(n)=\gamma$$
It numerical value is $\gamma=0.5772166...$
Let B be valid for all integers, where $B\ge 2$
Show that,
$$\int_0^1\left[\frac{B-1}{1-x}+\frac{B(x^B-x)}{(1-x)(1-x^B)}\right]\sum_{n=1}^{\infty}x^{B^n-1}dx=\gamma$$
I will be greatly appreciated if anyone can prove it for me.
 [/INST] Let we set $f_B(x)=\sum_{n\geq 1}x^{B^n}$. We have to compute:
$$ I_B=\int_{0}^{1}\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right)\,f_B(x)\,\frac{dx}{x} $$
but since $f_B(x^B) = x^B+f_B(x)$, by setting $x=z^B$ it follows that:
$$ I_B = \int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)\cdot(z^B+f_B(x))\cdot\frac{dz}{z}$$
and by using the digamma function we also have:
$$ \int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)z^{B-1}\,dz = -\gamma-\psi\left(\frac{1}{B}\right) $$
as well as:
$$ \int_{0}^{1}\left(\frac{B^{n+1}}{1-z^{B^n}}-\frac{B^n}{1-z^n}\right)z^{B^n-1}\,dz = -\gamma-\psi\left(\frac{1}{B}\right) $$
hence:
$$\begin{eqnarray*} I_B &=& -\gamma-\psi\left(\frac{1}{B}\right)+\int_{0}^{1}\left(\frac{B^2}{1-z^{B^2}}-\frac{B}{1-z^B}\right)f_B(z)\,\frac{dz}{z}\end{eqnarray*}$$
and the claim follows by using induction and the fact that, in a neighbourhood of zero:
$$ \psi(x)=-\frac{1}{x}-\gamma+\zeta(2)x+O(x^2). $$
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