Let $I_n = \int_0^1 x^ne^{-x} dx$, show that $0 < I_n < \frac{1}{n+1}$

Question

The question ask us to use the fact that if $f(x) < g(x)$, $\forall x \in [a,b]$, then $\int_a^b f(x) dx < \int_a^b g(x) dx $.

I have no idea about how to choose $f(x)$ so that $\int_0^1 f(x) dx$ will smaller than $\int_0^1 x^ne^{-x} dx$ (which is $I_n$) and $\int_0^1 f(x) dx$ will be $0$.

I have tried letting $f(x) = 0$ but it seems not working.

And I also don't know how to choose $g(x)$. I have tried to let $g(x) = \frac{-1}{(n+1)^2}$(becaues its integral will be $\frac{1}{n+1}$, but I don't know how to compare $\frac{-1}{(n+1)^2}$ with $ x^ne^{-x}$

Thanks for giving any tips in advance!

Answer 1

$e^{x}>1$ for $x\in(0,1]$ and hence $e^{-x}<1$ for all such $x$, now $\displaystyle\int_{0}^{1}x^{n}e^{-x}dx<\int_{0}^{1}x^{n}dx=\dfrac{1}{n+1}$.

Answer 2

You can do a little better.

By the integral Chebychev inequality (https://en.wikipedia.org/wiki/Chebyshev%27s_inequality#Integral_Chebyshev_inequality), since $x^n$ and $e^{-x}$ are monotonic in different directions,

$\int_0^1 x^ne^{-x} dx \le \int_0^1 x^n dx\int_0^1e^{-x} dx =\dfrac{1-1/e}{n+1} $.

Answer 3

An exact answer can be gotten by using the fact that this integral is the lower incomplete gamma function $\gamma(n+1, 1)$.

Since $\gamma(s, x) =\int_0^x t^{s-1}e^{-t}dt =x^s\sum_{k=0}^{\infty} \dfrac{(-x)^k}{k!(s+k)} $, we have $\gamma(n+1, 1) =\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(n+1+k)} $.

Since the sum is an alternating series with decreasing terms, its value is between any two consecutive sums.

The first terms are $\dfrac1{n+1}, -\dfrac1{1(n+2)}, \dfrac1{2(n+3)} $, so the sum is between $\dfrac1{n+1} -\dfrac1{(n+2)} =\dfrac{1}{(n+1)(n+2)} $ and $\dfrac1{n+1} -\dfrac1{1(n+2)} +\dfrac1{2(n+3)} =\dfrac{n^2 + 5 n + 8}{2 (n + 1) (n + 2) (n + 3)}\\ =\dfrac1{2n}-\dfrac{n^2 + 3 n + 6}{2 n (n + 1) (n + 2) (n + 3)} $.

Another formula which show the actual growth of the integral is

$\gamma(s, x) =x^s e^{-x}\sum_{k=0}^{\infty} \dfrac{x^k}{s(s+1)...(s+k)} $ so $\gamma(n+1, 1) = e^{-1}\sum_{k=0}^{\infty} \dfrac{1}{(n+1)(n+2)...(n+1+k)}\\ =\dfrac1{e(n+1)}+O(\dfrac1{n^2}). $

Answer 4

By integrating multiple times the convexity inequality $e^{-x}\geq 1-x$ we have that $g(x)=\frac{1-x-e^{-x}}{x^2}$ is bounded between $-\frac{1}{2}$ and $-\frac{1}{e}$ over $(0,1)$. It follows that

$$ I(n)=\int_{0}^{1}x^n e^{-x}\,dx = \int_{0}^{1}x^n(1-x-x^2 g(x))\,dx =\frac{1}{(n+1)(n+2)}+\frac{C}{n+3},\quad C\in\left[\frac{1}{e},\frac{1}{2}\right].$$

We may also notice that $$ (n+1) I(n) \stackrel{\text{IBP}}{=} \frac{1}{e}+I(n+1) $$ so by induction $$ \int_{0}^{1}x^n e^{-x}\,dx = \frac{1}{e(n+1)}+\frac{1}{e(n+1)(n+2)}+\frac{1}{e(n+1)(n+2)(n+3)}+\ldots $$ where the RHS is obviously less than $$ \sum_{k\geq 1}\frac{1}{e(n+1)^k} = \frac{1}{en}.$$