A $q$-Pochhammer integral: $\int_0^\infty\!\frac1{1+x}\frac2{2+x}\frac4{4+x}\frac8{8+x}\cdots\,dx=\log2$ Question: Show that for any $a>1$, we have
$$\int_0^\infty\!\frac{dx}{(1+x)(1+\frac xa)(1+\frac x{a^2})(1+\frac x{a^3})\cdots}\overset?=\log a.$$
WolframAlpha is able to evaluate specific instances of the above integral numerically, but not symbolically.
For fixed $a>1$, denote the integrand by $f_a(x)$. It is not hard to check that $f_a$ is characterised by the following properties:

*

*$f_a:[0,\infty)\to\mathbb R$ is continuous;

*$f_a(0)=1$;

*$(1+x)f_a(x)=f_a(\frac xa)$ for all $x>0$.

We also have other identities relating the integrands for different values of $a$, eg. $f_a(x)=f_{a^2}(x)f_{a^2}(\frac xa)$.
In fact, we can write $f_a(x)$ in terms of the q-Pochhammer symbol, namely
$$f_a(x)=\frac1{\left(-x;\frac1a\right)_\infty}.$$
It is not clear to me that the desired identity is an easy consequence of above functional equations and relations. It is also not obvious how to set up the usual tools for 1D definite integrals (Feynman trick, complex contour integration, ...).
 A: For $n \in \mathbb{N}$ and $a > 1$ let
$$ I_n (a) = \int \limits_0^\infty \frac{\mathrm{d} x}{\prod \limits_{k=0}^n (1 + x/a^k)} \, . $$
Partial fractions lead to
\begin{align*}
I_n (a) &= \int \limits_0^\infty \left[\sum \limits_{k=0}^n \left(\prod \limits_{k \neq l = 0}^n (1 - a^{k-l})^{-1} \right) \frac{1}{1+x/a^k}\right] \mathrm{d} x \\
&= \lim_{r \to \infty} \sum \limits_{k=0}^n a^k \left(\prod \limits_{k \neq l = 0}^n (1 - a^{k-l})^{-1} \right) \log(1+r/a^k) \\
&= \lim_{r \to \infty} \sum \limits_{k=0}^n a^k \left(\prod \limits_{k \neq l = 0}^n (1 - a^{k-l})^{-1} \right) [\log(r) - k \log(a)] \, .
\end{align*}
Now we rewrite
\begin{align}
a^k \prod \limits_{k \neq l = 0}^n (1 - a^{k-l})^{-1} &= \left(\prod \limits_{m=1}^n (1-a^{-m})^{-1}\right) (-1)^k a^{-k(k-1)/2} \prod \limits_{m=1}^k \frac{1 - a^{-(n-m+1)}}{1 - a^{-m}} \\
&= \left(\prod \limits_{m=1}^n (1-a^{-m})^{-1}\right) (-1)^k a^{-k(k-1)/2} \binom{n}{k}_{1/a}
\end{align}
in terms of the Gaussian binomial coefficient. The corresponding binomial theorem allows us to compute the sum over $k$:
\begin{align}
I_n(a) &= \left(\prod \limits_{m=1}^n (1-a^{-m})^{-1}\right) \lim_{r \to \infty} \sum \limits_{k=0}^n [\log(r) - k \log(a)] (-b)^k a^{-k(k-1)/2} \binom{n}{k}_{1/a} \, \Bigg \vert_{b=1} \\
&= \left(\prod \limits_{m=1}^n (1-a^{-m})^{-1}\right) \lim_{r \to \infty} \left[\log(r) - \log(a) \frac{\mathrm{d}}{\mathrm{d}b}\right] \prod \limits_{k=0}^{n-1} (1 - b a^{-k}) \, \Bigg \vert_{b=1} \\
&= \left(\prod \limits_{m=1}^n (1-a^{-m})^{-1}\right) \log(a) \prod \limits_{k=1}^{n-1} (1 - a^{-k}) \\
&= \frac{\log(a)}{1-a^{-n}} \, .
\end{align}
For your integral the dominated convergence theorem implies
$$ \int \limits_0^\infty \frac{\mathrm{d} x}{\prod \limits_{k=0}^\infty (1 + x/a^k)} = \lim_{n \to \infty} I_n (a) = \log(a) $$
as conjectured.

Given the simple result for $I_n (a)$, I had hoped for a more 'straightforward' proof. Some ideas:

*

*A contour integral approach yields $$ I_n(a) = \sum \limits_{k=0}^n \operatorname{Res} \left(z \mapsto \frac{-\log(-z)}{\prod \limits_{l=0}^n (1+z/a^l)},-a^k\right) \, , $$ which is clearly equivalent to the partial fractions method.


*Using Feynman parametrisation, we can derive $$ I_n (a) = (n-1)! \int \limits_{\Delta_n} \frac{\mathrm{d}u}{1 - \sum \limits_{k=1}^n (1-a^{-k}) u_k} \, , $$ where $\Delta_n = \{u \in [0,1]^n : \sum_{k=1}^n u_k \leq 1\}$. I have not been able to compute the remaining integral for arbitrary $n$.
