equivalent of an integral I am looking for an equivalent of the following integral:
$$\int_{n}^{+\infty}\log(1-2^{-t})\mathrm dt$$ when $n\to+\infty$.
Any hint or solution will be welcome. Thanks in advance
 A: If by "equivalent" you mean asymptotic expansion, then
$$
\log(1-2^{-t})\approx-2^{-t}-\tfrac122^{-2t}-\tfrac132^{-3t}-\tfrac142^{-4t}-\dots
$$
so
$$
\int_n^\infty\log(1-2^{-t})\,\mathrm{d}t\approx-\tfrac1{\log(2)}2^{-n}-\tfrac1{4\log(2)}2^{-2n}-\tfrac1{9\log(2)}2^{-3n}-\tfrac1{16\log(2)}2^{-4n}-\dots
$$
A: Your integral has an analytic expression:
$$
 \int \ln \left( 1 - 2^{-t} \right) \, dx= \frac{1}{\ln (4)} \left(
t \ln (2) \left(t \ln (2)+2 \ln \left(1-2^{-t}\right)-2 \ln \left(1-2^t\right)\right)-2 \text{Li}_2\left(2^t\right)
\right)
$$
where $\text{Li}_2(t)$ is the polylogarithm function.
$$
\int_0^{\infty } \ln \left(1-2^{-t}\right) \, dt = -\frac{\pi^{2}}{\ln (64)}
\approx -2.373138220831250905643446
$$

Asymptotic behavior:
$$
 \lim_{t\to\infty} \left( 1 - 2^{-t} \right) = 1
$$
Therefore
$$
\lim_{t\to\infty} \ln \left( 1 - 2^{-t} \right) = 0
$$
The function is plotted below.



Thanks to @user58697 for pointing out the left hand boundary point is not fixed.
$$
\int_x^{\infty } \ln \left(1-2^{-t}\right) \, dt =
2 \frac{\text{Li}_2\left(2^x\right)}{\ln (4)}- x^2 \frac{ \ln (2)}{2} 
+x \ln \frac{ 1-2^x }{1-2^{-x}}
-\frac{2 \pi ^2}{\ln (64)}
$$
