# Find $\lim_{n \rightarrow \infty} \int_{0}^{x} \frac{(-\ln{z})^n}{n!} dz$

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?

• notice that $n! = o(n^n)$ – user12986714 May 2 '20 at 3:05
• still , does that make the limit calculation easy? – Maths Freak May 2 '20 at 3:23
• I found a way of using gamma function. The answer is 1 – Isaac YIU Math Studio May 2 '20 at 3:24
• yes, please share your approach – Maths Freak May 2 '20 at 3:25
• @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: $$1-\lim_{n \to \infty} \dfrac{1}{n!} \int ^a_0 u^n e^{-u} du$$ – 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

$$\lim_{n\to\infty}f_n(x)=1$$

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

Not a direct proof but one can note that $$f(z)=\frac{(-\ln z)^n}{n!}1_{0 is the density function of $$\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

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

Hence,

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

• 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. – StubbornAtom May 2 '20 at 7:17
• The Gamma-Poisson relationship is derived here and here for example, as @AdityaSriram has shown. – StubbornAtom May 2 '20 at 8:06