Find $\lim_{n \rightarrow \infty} \int_{0}^{x} \frac{(-\ln{z})^n}{n!} dz$ where $x \in (0,1)$ .

I was thinking of using Dominated Convergence Theorem to the function $f_n(z)=\frac{(-\ln{z})^n}{n!} dz$ But couldn't find its limiting function. Any other approach which can help?

  • $\begingroup$ notice that $n! = o(n^n) $ $\endgroup$ – user12986714 May 2 '20 at 3:05
  • $\begingroup$ still , does that make the limit calculation easy? $\endgroup$ – Maths Freak May 2 '20 at 3:23
  • $\begingroup$ I found a way of using gamma function. The answer is 1 $\endgroup$ – Isaac YIU Math Studio May 2 '20 at 3:24
  • $\begingroup$ yes, please share your approach $\endgroup$ – Maths Freak May 2 '20 at 3:25
  • $\begingroup$ @MathsFreak I am not quite sure, as it is checked from wolfram alpha. I substitute $u=-\ln z$ and $a=-\ln x$, and it turns out to have: $$ 1-\lim_{n \to \infty} \dfrac{1}{n!} \int ^a_0 u^n e^{-u} du$$ $\endgroup$ – Isaac YIU Math Studio May 2 '20 at 3:27

Let us first consider the integral, $${I(n)} = \int_0^{x}\frac{(-\ln z)^n}{n!}dz$$ Using Integration by Parts, $$I(n) = \frac{z(-\ln z)^n}{n!}|_0^{x}+n\int_0^{x}\frac{(-\ln z)^{n-1}}{n!}dz=\frac{x(-\ln x)^n}{n!}+I(n-1)$$ If you observe, the above is a telescopic sum, $$\therefore I(n)=I(0)+x\sum_{i=1}^{n}\frac{(-\ln x)^i}{i!}$$ Since $I(0) =x$, $$I(n)=x\sum_{i=0}^{n}\frac{(-\ln x)^i}{i!}$$ As $n \to \infty$, $$\lim_{n \to \infty}I(n)=x\sum_{i=0}^{\infty}\frac{(-\ln x)^i}{i! }=x\exp{(-\ln x)}=1$$


Enforcing the substitution $z\mapsto e^{-z}$ and recalling that $n!=\Gamma(n+1)=\int_0^\infty z^ne^{-z}\,dz$ we can write

$$\begin{align} f_n(x)&=\int_0^x\frac{(-\log(z))^n}{n!}\,dz\\\\ &=\frac{1}{n!}\int_{\log(1/x)}^\infty z^ne^{-z}\,dz\\\\ &=\frac1{n!}\int_0^\infty z^ne^{-z}\,dz-\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz\\\\ &=1-\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz \end{align}$$

Finally, using the estimate

$$\left|\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz\right|\le \frac{\log^{n+1}(1/x)}{n!} $$

the squeeze theorem guarantees that $\lim_{n\to \infty }\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz=0$.

Putting it all together, yields the coveted limit


for $x\in (0,1)$. And we are done!


Not a direct proof but one can note that $$f(z)=\frac{(-\ln z)^n}{n!}1_{0<z<1}$$ is the density function of $\prod\limits_{i=1}^{n+1} X_i$ where $X_1,\ldots,X_{n+1}$ are i.i.d uniform on $(0,1)$.

So for $x\in (0,1)$,

\begin{align} \int_0^x f(z)\,dz&=P\left(\prod_{i=1}^{n+1} X_i \le x\right) \\&=P\left(\sum_{i=1}^{n+1} \ln X_i \le \ln x\right) \end{align}

Now $-\ln X_i$'s are i.i.d exponential with mean $1$, so by classical CLT

$$\sqrt{n+1}\left(\frac1{n+1}\sum_{i=1}^{n+1} \ln X_i +1\right)\stackrel{L}\longrightarrow N(0,1)$$


$$ P\left(\sum_{i=1}^{n+1} \ln X_i \le \ln x\right)\approx \Phi\left(\frac{\ln x}{\sqrt{n+1}}+\sqrt{n+1}\right) \stackrel{n\to\infty}\longrightarrow \Phi(\infty)=1 $$

  • $\begingroup$ Actually, we have by the relation of Poisson and Gamma distribution that $$\int_0^x \frac{(-\ln z)^n}{n!}\,dz=1-\int_0^{-\ln x}\frac{e^{-u}u^n}{n!}\,du=\sum_{j=0}^n \frac{e^{\ln x}(-\ln x)^j}{j!} $$ As $n\to \infty$, that last sum obviously converges to $1$ as it is a PMF. $\endgroup$ – StubbornAtom May 2 '20 at 7:17
  • $\begingroup$ The Gamma-Poisson relationship is derived here and here for example, as @AdityaSriram has shown. $\endgroup$ – StubbornAtom May 2 '20 at 8:06

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