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Evaluate $\displaystyle \lim_{m\rightarrow \infty}\bigg[m^3\int^{2m}_{m}\frac{xdx}{x^5+1}\bigg]$ for $m\in\mathbb{N}$

what i try

put $x^5+1=t$ and $dx=\frac{1}{5}x^{-4}dt=\frac{1}{5x^4}dt$

$$\lim_{n\rightarrow \infty}\frac{m^3}{5}\bigg[\int^{32m^5+1}_{m^5+1}\frac{x^2}{t(t-1)}dt\bigg]$$

How do i solve it. Help me please

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  • 3
    $\begingroup$ Hint: Apply L'Hopital's rule. $\endgroup$ – Paras Khosla Mar 26 '19 at 7:32
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We have $\int^{2m}_{m}\frac{xdx}{x^5+1} \leq m\cdot \frac{2m}{m^5+1} \stackrel{m\to \infty}{\longrightarrow} 0$

Set $I(m) = \int^{2m}_{m}\frac{xdx}{x^5+1} \Rightarrow m^3\cdot I(m) = \frac{I(m)}{\frac{1}{m^3}}$.

So, we have a L'Hospital case of $\frac{0}{0}$:

\begin{eqnarray*} \frac{I(m)}{\frac{1}{m^3}} & \stackrel{L'Hospt.}{\sim} & \frac{2\frac{2m}{(2m)^5+1}-\frac{m}{m^5+1}}{-3\cdot m^{-4}} \\ & = & -\frac{1}{3}\left(\frac{4m^5}{32m^5+1} - \frac{m^5}{m^5+1}\right) \\ & \stackrel{m \to \infty}{\longrightarrow} & \frac{7}{24} \end{eqnarray*}

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Putting $x=mt$ in the integral we see that the desired limit is equal to the limit of the expression $$m^5\int_{1}^{2}\frac{t\,dt}{m^5t^5+1}$$ which is same as the limit $$\lim_{h\to 0^{+} }\int_{1}^{2}\frac{t\,dt}{h+t^5}$$ via the substitution $h=1/m^5$. Clearly the integrand is continuous in both $t, h$ and thus the limit can be taken inside to get the desired limit as $\int_{1}^{2}t^{-4}\,dt=7/24$.


Another simpler approach is to note that $$m^3\int_m^{2m}\frac{dx}{x^4}=\frac{7}{24}$$ and we have therefore $$\left|m^3\int_m^{2m}\frac{x\,dx}{1+x^5}-\frac{7}{24}\right|=m^3\int_{m}^{2m}\frac{dx}{x^4(1+x^5)}\leq m^3\cdot m\cdot\frac{1}{m^4(1+m^5)}$$ and the rightmost expression tends to $0$. This avoids the interchange of limit with integral and is fully within the scope of a typical high school calculus course.

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  • 2
    $\begingroup$ Nice way to do it ! Using it, and using series, we could find the asymptotics $\frac{7}{24}-\frac{255}{2048 m^5}+O\left(\frac{1}{m^{10}}\right)$ $\endgroup$ – Claude Leibovici Mar 26 '19 at 9:43
  • $\begingroup$ @ClaudeLeibovici: I have obtained a rather crude approximation of order $O(m^{-5})$ in later part of updated answer. $\endgroup$ – Paramanand Singh Mar 26 '19 at 10:43
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Another intuive but non-rigorous argument. Intuitively as the lower bound of the integral in the denominator, $m\to \infty$ so the integrand goes to $0$ so we can say essentially the integral in the denominator goes to $0$. This is very inviting for an application of L'Hopital's rule which gives you $7/24$ as the limiting value as follows. We can approximate the integral by: $$\begin{aligned}\int_{m}^{2m}\dfrac{1}{x^4}\mathrm dx=\dfrac{-1}{3(2m)^3}+\dfrac{1}{3m^3}\implies \lim_{m\to \infty}m^3\int_{m}^{2m}\dfrac{x}{x^5+1}\mathrm dx\to \dfrac{-1}{24}+\dfrac{1}{3}=\boxed{\dfrac{7}{24}}\end{aligned}$$

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  • $\begingroup$ +1 Nice. ${}{}{}{}{}{}{}{}$ $\endgroup$ – Felix Marin Mar 27 '19 at 23:50

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