Bound this integral $\displaystyle\int_0^\infty \frac{(1-p+pe^{-t})^n -e^{-npt} -(1-p)^n(1-e^{-nt})}{1-e^{-t}}dt$

Question

If $n\in\mathbb N$ and $p\in (0,1)$, can the integral below be bounded by $\frac{1}{np}$ or $\frac{1}{np} +\mathcal O(n^{-1})$? $$\int_0^\infty \frac{(1-p+pe^{-t})^n -e^{-npt} -(1-p)^n(1-e^{-nt})}{1-e^{-t}}dt$$

Answer 1

Definitely, there should be a simple way to find the asymptotics of this integral (the answer is simple), but I did not manage to find it.

So, let's take the substitution $e^{-t}=x$ $$I(p)=\int_0^1\frac{dx}{x(1-x)}\Big((1-p+px)^n-x^{np}-(1-p)^n(1-x)^n\Big)$$ The integrand is well-defined at $x=0$ and $x=1$. It is also useful to note that $I(p=1)=0$. Using $\frac{1}{x(1-x)}=\frac{1}{x}+\frac{1}{1-x}$ , and making the substitution $t=1-x$ in the second term (to replace $\frac{1}{1-x}$ by $\frac{1}{t}$), the integral gets the form $$I=\int_0^1\frac{dx}{x}\Big((1-p+px)^n-x^{np}-(1-p)^n(1-x)^n+(1-px)^n-(1-x)^{np}-(1-p)^n(1-(1-x)^n)\Big)$$ Integrating by part ($\,\frac{dx}{x}=d(\ln x)\,$) $$I=-pn\int_0^1\ln x(1-p+px)^{n-1}dx+pn\int_0^1\ln x\, x^{np-1}dx-n(1-p)^n\int_0^1\ln x \,x^{n-1}dx$$ $$+pn\int_0^1\ln x (1-px)^{n-1}dx-pn\int_0^1\ln x\, (1-x)^{np-1}dx+n(1-p)^n\int_0^1\ln x \,(1-x)^{n-1}dx$$

$$=I_1+I_2+I_3+I_4+I_5+I_6$$ Evaluating every integral, we get $$I_2=pn\int_0^1\ln x\, x^{np-1}dx=-\frac{1}{np}$$ $$I_3=-n(1-p)^n\int_0^1\ln x\, x^{n-1}dx=-\frac{(1-p)^n}{n}$$ $$I_5=-np\int_0^1\ln x\, (1-x)^{np-1}dx=-np\frac{\partial}{\partial s}\int_0^1x^s(1-x)^{np-1}dx\Big|_{s=0}$$ $$=-np\frac{\partial}{\partial s}B(1+s;np)\Big|_{s=0}=-np\Big(\psi(1)-\psi(1+np)\Big)$$ where $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ - digamma-function $$I_6=n(1-p)^n\int_0^1\ln x\,(1-x)^{n-1}dx=(1-p)^n\Big(\psi(1)-\psi(1+n)\Big)$$ $I_1$ and $I_4$ are more complicated - we will find their asymptotics at $n\to\infty$ Making the substitution $t=1-x$ $$I_1=-np\int_0^1\ln(1-x)(1-px)^{n-1}dx=-np\int_0^1\ln(1-x)e^{(n-1)\ln(1-px)}dx$$ $$=-np\int_0^1\ln(1-x)e^{-p(n-1)x}\Big(1-\frac{n-1}{2}p^2x^2+...\Big)dx$$ Making the substitution $p(n-1)x=t$ and supposing that $p(n-1)\gg1\,$ (it means also that $p$ cannot be too close to zero) $$I_1\sim-\frac{np}{p(n-1)}\int_0^{p(n-1)}\ln\Big(1-\frac{t}{p(n-1)}\Big)e^{-t}dt$$ $$+\frac{n(n-1)p^3}{2}\frac{1}{p^3(n-1)^3}\int_0^{p(n-1)}\ln\Big(1-\frac{t}{p(n-1)}\Big)e^{-t}t^2dt$$ The main contribution to the integral comes from the region $t\ll p(n-1)$ (at $t\sim p(n-1)$ the exponent becomes very small). Using $\ln\Big(1-\frac{t}{p(n-1)}\Big)\approx \frac{t}{p(n-1)}$ and extending the integration to $\infty$ , we get (keeping only the terms $\sim \frac{1}{n}$ - the second integral, therefore, does not contribute) $$I_1=\frac{n}{p(n-1)^2}+O\Big(\frac{1}{n^2}\Big)=\frac{1}{pn}+O\Big(\frac{1}{n^2}\Big)$$ Using the same approach to $I_4$ , we get $$I_4\sim\frac{pn}{p(n-1)}\int_0^{p(n-1)}\ln\frac{t}{p(n-1)}e^{-t}dt-\frac{n(n-1)p^3}{2}\frac{1}{(n-1)^3p^3}{2}\int_0^{p(n-1)}\ln\frac{t}{p(n-1)}e^{-t}t^2\,dt$$ The second term is important and we have to keep it. Extending integration to $\infty$ and using $\int_0^\infty\int e^{-t}dt=\psi(1)=-\gamma$ $$I_4\sim\frac{n}{n-1}\Big(\psi(1)-\ln\big(p(n-1)\big)\Big)$$ $$-\,\frac{n}{2(n-1)^2}\bigg(\int_0^\infty\ln t \,e^{-t}t^2\,dt-2\ln\big(p(n-1)\big)\bigg)$$ Given that $\int_0^\infty\ln t \,e^{-t}t^2\,dt=\psi(3)\Gamma(3)=3+2\psi(1)$ $$I_4=\frac{n(n-2)}{(n-1)^2}\psi(1)-\frac{n(n-2)}{(n-1)^2}\ln\big(p(n-1)\big)-\frac{3}{2}\frac{n}{(n-1)^2}+O\Big(\frac{1}{n^2}\Big)$$ Using $\frac{n(n-2)}{(n-1)^2}=1+O\Big(\frac{1}{n^2}\Big)$, with the required accuracy, $$I_4=\psi(1)-\ln\big(p(n-1)\big)-\frac{3}{2n}+O\Big(\frac{1}{n^2}\Big)$$ Now, gathering all together and using $\psi(1+np)=\ln (np)+\frac{1}{2np}+O\Big(\frac{1}{n^2}\Big)$

$$I(p)=\frac{1}{np}-\frac{1}{np}+\frac{(1-p)^n}{n}+\psi(1)-\ln\big(p(n-1)\big)-\frac{3}{2n}+\ln(np)+\frac{1}{2np}-\psi(1)+(1-p)^n\Big(\psi(1)-\psi(1+n)\Big)+O\Big(\frac{1}{n^2}\Big)$$ Given the requirement $p(n-1)\gg1$ we are not allowed to keep the terms$\sim(1-p)^n\sim e^{n\ln(1-p)}\sim e^{-np}\ll\frac{1}{n}$. Also, $\,\ln(n-1)=\ln n-\frac{1}{n}+O\Big(\frac{1}{n^2}\Big)$.

Therefore, with the required accuracy, $$\boxed{\,\,I(p)\sim\frac{1}{2pn}-\frac{1}{2n}=\frac{1-p}{2pn}\,;\,\,np\gg1\,\,}$$

We see that at $p=1\,\,I(p)=0\,\,$ - as required.

Quick check: WolframAlpha gives, for example: $$p=0.9\,\, n= 20 \,\,I=0.00305...\,\, (\text{asymptotics} - 0.002777... )$$ $$p=0.5\,\, n= 100 \,\,I=0.0005181...\,\, (\text{asymptotics} - 0.0005 )$$

$\mathbf{Addendum:}$ In the same way the second term of the asymptotics can be evaluated $$\boxed{\,\,I(p)\sim\frac{1-p}{2pn}+\frac{1}{(pn)^2}\Big(\frac{5}{12}(1-p^2)+p(1-p)\Big)\,;\,\,np\gg1\,\,}$$ $$I(1)=0$$ The accuracy now is $\,\sim 1.5$%. Quick check:

$$p=0.5\,\, n= 100 \,\,I=0.0005181...\,\, (\text{asymptotics} - 0.000525 )$$

Answer 2

Using the fact that $\frac{1}{1-e^{-t}} = \sum_{k=0}^\infty e^{-kt}$ on the first two terms of the numerator and using $1-e^{-nt} = 1^n-e^{-nt} = (1-e^{-t}) \sum_{k=0}^{n-1} e^{-kt}$, we have \begin{align} &\int_0^\infty \frac{(1-p+pe^{-t})^n -e^{-npt} -(1-p)^n(1-e^{-nt})}{1-e^{-t}}dt \\ &= \int_0^\infty \sum_{k=0}^\infty e^{-kt} \left[(1-p+pe^{-t})^n -e^{-npt}\right] -\sum_{k=0}^{n -1} e^{-kt} (1-p)^n dt\\ &= \sum_{k=0}^{n-1} \int_0^\infty e^{-kt} \left[(1-p+pe^{-t})^n -e^{-npt} -(1-p)^n\right] dt \\ &\quad +\sum_{k=n}^\infty \int_0^\infty e^{-kt} \left[(1-p+pe^{-t})^n -e^{-npt}\right] dt \end{align}

Using the binomial theorem and $\int_0^\infty e^{-at}dt = \frac{1}{a}$ on the first summation \begin{align} &\sum_{k=0}^{n-1} \int_0^\infty e^{-kt} \left[(1-p+pe^{-t})^n -e^{-npt} -(1-p)^n \right] dt \\ &= \sum_{k=0}^{n-1} \int_0^\infty e^{-kt} \left[\sum_{m=1}^n e^{-mt} \binom{n}{m}p^m(1-p)^ {n-m} -e^{-npt}\right] dt \\ &= \sum_{k=0}^{n-1} \int_0^\infty \left(\sum_{m=1}^n e^{-(m+k)t} \binom{n}{m}p^m(1-p)^{n- m} -e^{-(np+k)t} \right) dt \\ &= \sum_{k=0}^{n-1} \sum_{m=1}^n \frac{1}{m+k} \binom{n}{m}p^m(1-p)^{n-m} -\frac{1} {np+k} \end{align} I'm not finished this part yet but I think I will try to bound the inner summation.

For the second summation, I'm using $\left|a^n-b^n \right| \leq (\max\{a,b\})^{n-1} n|a-b|$ then using a Taylor expansion of the exponentials because the first 2 terms cancel, so with the alternating series bound, we get this:

\begin{align} &\left| \sum_{k=n}^\infty \int_0^\infty e^{-kt} \left[(1-p+pe^{-t})^n -e^{-npt}\right] dt \right| \\ &\leq \sum_{k=n}^\infty \int_0^\infty e^{-kt} (1-p+pe^{-t})^{n-1} \frac{np(1-p)t^2}{2} dt \\ &= \sum_{k=n}^\infty \int_0^\infty \sum_{m=0}^{n-1} \frac{np(1-p) e^{-(m+k)t} t^2}{2} \binom{n-1} {m}p^m(1-p)^{n-1-m} dt \\ &= \sum_{k=n}^\infty \sum_{m=0}^{n-1} \frac{np(1-p) \int_0^\infty e^{-(m+k)t} t^2dt}{2} \binom{n- 1}{m}p^m(1-p)^{n-1-m} \\ &= \sum_{k=n}^\infty \sum_{m=0}^{n-1} \frac{np(1-p) \Gamma(3)}{2(m+k)^2} \binom{n-1}{m}p^m(1-p)^{ n-1-m} \\ &= \sum_{k=0}^\infty \sum_{m=0}^{n-1} \frac{np(1-p)}{(n+m+k)^2} \binom{n-1}{m}p^m(1-p)^{n-1-m} \\ \end{align}

I'm still working on the last details of bounding the summations.