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The value of

$$\lim_{n\to \infty} \int_0^1 nx^n e^{x^2} dx$$

is ____________(round off to three decimal places)

I tried integrating by parts and bring out some recurrence relation , but it was of no use.

However, I can show the limit to be greater than or equal to 1.

$\int_0^1 nx^n e^{x^2} dx \gt \int_0^1 nx^{n+2} dx$ (using $e^x\gt x $ for $x\gt 0$)

$=\frac n{n+3}=1-\frac 3{n+3}$

Taking limit as $n\to \infty$ , proves my assertion.

Please help me find the actual limit.Thanks in advance.

StubbornAtom
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user-492177
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4 Answers4

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Put $y=x^{n}$. The integral becomes $\int_0^{1} y^{1/n}e^{y^{2/n}}dy$. Check that the integrand is bounded by $e$ and that it converges point-wise to $e$. By DCT the answer is $e$.

Kavi Rama Murthy
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Alternatively, \begin{equation} \int_0^1 nx^n e^{x^2} dx = \int_0^1nx^n\sum_{k=0}^\infty \frac{x^{2k}}{k!} = \sum_{k=0}^\infty \int_0^1\frac{nx^{2k+n}}{k!} = \sum_{k=0}^\infty \frac{n}{(2k+n+1)k!}. \end{equation} Now, \begin{equation} \lim_{n\to\infty}\int_0^1 nx^n e^{x^2} dx = \sum_{k=0}^\infty\lim_{n\to\infty}\frac{n}{(2k+n+1)k!}= \sum_{k=0}^\infty\frac{1}{k!}=e \end{equation}

Revzora
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  • ok, but one question, what makes the interchange of sum and limit possible? – user-492177 Jun 04 '20 at 11:02
  • You can show https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged. If you don't already know the Lebesgue integral, you can only take the statments. – Revzora Jun 04 '20 at 11:19
  • However, even in mathematics, it is sometimes important not to be too strict. – Revzora Jun 04 '20 at 11:25
  • @Revzora..Yes, I agree, but I am quite an extremist , hence the question . Anyway, thanks for your alternative answer. – user-492177 Jun 04 '20 at 11:37
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$$I(n)=\int_0^1nx^ne^{x^2}dx$$ $u=x^2$ then $dx=\frac{du}{2x}$ and so: $$I(n)=\frac n2\int_0^1x^{n-1}e^{x^2}du=\frac n2\int_0^1u^{\frac{n-1}{2}}e^udu$$ now: $$e^u=\sum_{k=0}^\infty\frac{u^k}{k!}$$ and so: $$I(n)=\frac{n}{2}\int_0^1\sum_{k=0}^\infty\frac{u^{\frac{2k+n-1}{2}}}{k!}du=n\sum_{k=0}^\infty\frac{1}{(2k+n-1)k!}$$ now notice that: $$\lim_{n\to\infty}I(n)=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{n}{(2k+n-1)k!}=\sum_{k=0}^\infty\frac{1}{k!}=e$$

Henry Lee
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Late answer because the question might be tagged as duplicate where I put this answer before:

Partial integration gives

$$I_n :=\int_0^1 \underbrace{nx^{n-1}}_{u'}\cdot\underbrace{xe^{x^2}}_{v}dx= \left.x^{n+1}e^{x^2}\right|_0^1- \underbrace{\int_0^1 x^n(1+2x^2)e^{x^2}dx}_{J_n=}=e-J_n$$

Now, $J_n$ can be easily estimated as follows $$0\leq J_n \leq 3e\int_0^1x^ndx=\frac{3e}{n+1}\stackrel{n\to \infty}{\longrightarrow}0$$

Hence, $I_n \stackrel{n\to \infty}{\longrightarrow} e$.

trancelocation
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