Any ideas on bounding $\int_{0}^\infty (1-(1-e^{-x})^d) dx$ in terms of $d$? Let $\epsilon_i(j)$ be i.i.d and take values in $[-1,1]$. The integral comes from a sum that arises as follows: Every turn $n$ we select an element of $\{0,1,2,\ldots, d\}$ as the least component of the vector
$$\left( 0,n + \sum_{i=1}^n \epsilon_i(1), n + \sum_{i=1}^n \epsilon_i(2) , \ldots, n + \sum_{i=1}^n \epsilon_i(d) \right).$$
If we select $d\ne 0$ the cost for that turn is $1$. Otherwise the cost is free. I am interested in bounding the expected cost.
Using concentration inequalities I can bound the probability of choosing $d \ne 0$ on turn $n$ as $(1-(1-e^{-Cn})^d)$  for some $C>0$. The expected cost overall is then $\sum_{n=1}^\infty(1-(1-e^{-Cn})^d)$ which has the same order as the integral $\int_{0}^\infty (1-(1-e^{-Cx})^d) dx$.
Feeding the integral to Wolfram Alpha we get something in terms of the Hypergeometric function. In any case there seems to be no elementary antiderivative. 
Numerically (
https://www.desmos.com/calculator/adndsq4le9) the sum seems to be about about $\frac{2.5}{C} \log (1+d)$. Any ideas on how we could prove some bound $ \sum_{n=0}^\infty(1-(1-e^{-Cn})^d) < \frac{D}{C} \log(d)$ for some $D>0$ say? 
If we can get a bound for $C=1$ we can do it for all $C$ since the subsstitution $y=Cx$ gives $ \int_{0}^\infty(1-(1-e^{-Cx})^d)dx =  \frac{1}{C}\int_{0}^\infty(1-(1-e^{-y})^d)dy$.
 A: For $d \in \mathbb{N}$ let $I(d) = \int_0^\infty [1 - (1-\mathrm{e}^{-x})^d]\, \mathrm{d} x$. The substitution $\mathrm{e}^{-x} = t$ yields
$$ I(d) = \int \limits_0^1 \frac{1 - (1-t)^d}{t} \, \mathrm{d} t = \int \limits_0^1 \frac{1 - u^d}{1-u} \, \mathrm{d} u = H_d $$
with the $d$-th harmonic number $H_d$. According to this paper, an optimal and a slightly weaker upper bound are given by
$$ I(d) = H_d < \ln(d) + \gamma + \frac{1}{2d + \frac{1}{3}} < \ln(d) + \gamma + \frac{1}{2d} \, . $$
A: Is it helpful to consider the binomial expansion $$(1-e^{-x})^d = \sum\limits_{k=0}^{d}{d\choose k}1^{d-k}(-e^{-x})^k=\sum\limits_{k=0}^{d}{d\choose k}(-1)^ke^{-kx}$$ since $d$ is integral? Then you would have
$$\int\limits_0^{\infty}(1-(1-e^{-x})^d)\;dx = \int\limits_0^{\infty}\left(1-\sum\limits_{k=0}^{d}{d\choose k}(-1)^ke^{-kx}\right)\;dx$$
$$= \int\limits_0^{\infty}\left(1-(1+\sum\limits_{k=1}^{d}{d\choose k}(-1)^ke^{-kx})\right)\;dx$$
$$= -\sum\limits_{k=1}^{d}{d\choose k}(-1)^k\int\limits_0^{\infty}e^{-kx}\;dx$$
$$=-\sum\limits_{k=1}^{d}{d\choose k}\frac{(-1)^k}{k}$$
which might be useful.
A: Playing around with bounds, let's try to prove that
$$1-(1-x)^d < d(xe^x)$$
for $0<x<1$ and assuming $d$ is an integer greater than $0$. 
Taking derivatives we have to compare $d(1-x)^{d-1}$ and $de^x(x+1)$. At $x=0$ the derivatives are equivalent but the function on the left is monotonically decreasing on $(0,1)$ whereas the function on the right is monotonically increasing since their derivatives (the original functions' second derivatives) are strictly negative and positive, respectively. Thus we have that
$$d(1-x)^{d-1} < de^x(x+1)$$
And since the original two functions equal each other at $x=0$, we can use the following theorem. If $f(a)=g(a)$ then
$$g'(x) < f'(x) \implies g(x) < f(x)$$
on $(a,b)$. Thus the desired inequality is proved taking the interval to be $(0,1)$.
Applying the inequality to the integral,
$$\int_0^\infty 1 - (1-e^{-x})^d dx < \int_0^\infty d\cdot e^{-x}e^{e^{-x}}dx = -d\cdot e^{e^{-x}}\Bigr|_0^\infty = d(e-1)$$
