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Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.

Hagen von Eitzen
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Hagrid
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4 Answers4

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Following the Ishan Banerjee comment let $t=x^n$ hence $dx=\frac{1}{n}t^{\frac{1}{n}-1}dt$ and then $$ (2n+1) \int_0 ^{1} x^n e^x dx=\frac{2n+1}{n}\int_0^1t^{1/n}e^{t^{1/n}}dt\to2e$$ by using the dominated convergence theorem.

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Let $I_n=\int\limits_0^1x^n\mathrm e^x\mathrm dx$. By integration by parts, $(n+1)I_n=\left.x^{n+1}\mathrm e^x\right|_0^1-I_{n+1}=\mathrm e-I_{n+1}$. Now, $0\leqslant I_{n+1}\leqslant I_n$ hence $(n+1)I_n\leqslant\mathrm e\leqslant(n+2)I_n$.

This is enough to show that $$ \left(2-\frac3n\right)\cdot\mathrm e\leqslant(2n+1)I_n\leqslant2\mathrm e, $$ hence $(2n+1)I_n\to2\mathrm e$.

Did
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A related technique. You can use integration by parts technique by letting $u=e^{x}$ which leads to

$$ I_n = \left( 2\,n+1 \right) \left( {\frac {{{\rm e}}}{n+1}}-{\frac {{ {\rm e}}}{2+3\,n+{n}^{2}}}+\int _{0}^{1}\!{\frac {{x}^{n+2}{ {\rm e}^{x}}}{ \left( n+2 \right) \left( n+1 \right) }}{dx} \right) .$$

$$\implies \lim_{n\to \infty}I_n = 2 \,\rm{e} + 0 + \lim_{n\to \infty } \int _{0}^{1}\!{\frac {(2n+1){x}^{n+2}{ {\rm e}^{x}}}{ \left( n+2 \right) \left( n+1 \right) }}{dx} $$

$$ \implies \lim_{ n\to \infty } = 2 \rm e . $$

Note that, the change of the limit with integral is due to the uniform convergence of the sequence $$ \frac {(2n+1){x}^{n+2}}{ \left( n+2 \right) \left( n+1 \right) }. $$

Here is a technique for proving uniform convergence.

Added: Integration by parts,

$$ \int u\, dv = u\,v -\int v \,du. $$

So, in your case $ u = e^{x} $ and $ dv = x^n dx $.

Community
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Mhenni Benghorbal
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  • I've been trying to get to this result for quite a long time, but I still don't know how you came up with it. Could you elaborate on your answer? – Hagrid May 19 '13 at 14:03
  • You mean $2 e$ I take it. – Ron Gordon May 19 '13 at 14:49
  • I think it is just a typo. I see that what is in the parentheses goes to e, but we also have 2n+1, so all in all the limit is 2e. What I don't know is how to get $I_n = \left( 2,n+1 \right) \left( {\frac {{{\rm e}}}{n+1}}-{\frac {{ {\rm e}}}{2+3,n+{n}^{2}}}+\int _{0}^{1}!{\frac {{x}^{n+2}{ {\rm e}^{x}}}{ \left( n+2 \right) \left( n+1 \right) }}{dx} \right) .$ – Hagrid May 19 '13 at 15:17
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    @Hagrid: The limit is $2,e$. It is a typo. As I said you need to integrate by parts twice and in each case you assume $u=e^x$. – Mhenni Benghorbal May 19 '13 at 16:36
  • Is it this expansion enough? Don't I need $\int x^n e^x dx = (\sum _{k=0} ^{n} (-1)^{n-k} \frac{n!}{k!}x^k)e^x +C$? – Hagrid May 19 '13 at 17:19
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    @Hagrid: yes, it is enough. Just note this, you can take the limit under the integral sign, because of the uniform convergence of the sequence on the interval $x\in [0,1]$. – Mhenni Benghorbal May 19 '13 at 17:43
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    Note that, since $ 0 \leq x \leq 1$, then $ \Big|\frac{(2n+1)x^{n+2}}{(n+1)(n+2)}-0\Big| \leq \frac{(2n+1)}{(n+1)(n+2)}<\epsilon. $ – Mhenni Benghorbal May 19 '13 at 17:51
  • Could you write that again? – Hagrid May 19 '13 at 17:54
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    @Hagrid: The last comment demonstrates the proof of the uniform convergence of the seq. – Mhenni Benghorbal May 19 '13 at 17:57
  • I see that now. Thank you. – Hagrid May 19 '13 at 18:00
  • @Hagrid: You are welcome. – Mhenni Benghorbal May 19 '13 at 18:06
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Here is an alternative argument. Note that for every $a\in [0,1)$, $$\frac{(1-a^{n+1})e^a}{n+1}=e^a\int_a^1x^ndx\le\int_a^1x^ne^xdx\le\int_0^1x^ne^xdx\le e \int_0^1x^ndx=\frac{e}{n+1}.$$ Multiplying the inequality above by $2n+1$ and letting $n\to\infty$, it follows that: $$2e^a\le\liminf_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le\limsup_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le 2e.$$ Since $a\in [0,1)$ is arbitrary, we can conclude that $$\lim_{n\to\infty}(2n+1)\int_0^1x^ne^xdx=2e.$$

Hu Zhengtang
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