Compute the following limit: $$\lim_{n\to\infty}\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}$$.

So I thought to use the result:

a) If $\{F_n\}$ converges uniformly on $S=[a,b]$ to $F$ and $F_n$ is integrable $\forall n$. Then $\int_\limits{a}^{b}F(x)dx=\lim\limits_{n\to\infty}\int_\limits{a}^{b}F_n(x)dx$.

$$ \lim_{n\to\infty}\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}=\lim_{n\to\infty}\int_\limits{0}^{1-\delta}\frac{dx}{x^{n}+1}+\lim_{n\to\infty}\int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1}+\lim_{n\to\infty}\int_\limits{1+\delta}^{\infty}\frac{dx}{x^{n}+1} $$

I can easily pass the limit under the integral in the first and third integral which gives me the results of $1-\delta$ and $0$ respectively.

However my problems lie in the second integral: $\lim\limits_{n\to\infty}\int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^n+1}$

I changed the variable $y=1-x$ which led me to $\lim\limits_{n\to\infty}\int_\limits{-\delta}^{\delta}\frac{dy}{(1-y)^{n}+1}=?$


How do I compute $\lim\limits_{n\to\infty}\int_\limits{-\delta}^{\delta}\frac{dy}{(1-y)^{n}+1}=$? As the delta is arbitrary I do not know how $1-y$ is going to behave.

  • 2
    $\begingroup$ Your last integral is bounded by $2\delta$, which is really all you need. $\endgroup$ – Lord Shark the Unknown Apr 22 '18 at 15:00
  • $\begingroup$ @LordSharktheUnknown I guess that is right, as $\delta\to 0$ the result would be one for the original expression. $\endgroup$ – Pedro Gomes Apr 22 '18 at 15:03
  • $\begingroup$ @LordSharktheUnknown I have been thinking about what you said. Is it true that the limit can go under the integral? $Y$ can be negative $\endgroup$ – Pedro Gomes Apr 22 '18 at 15:10
  • $\begingroup$ A super-overkill method is to evaluate the integral to $\frac{\pi/n}{\sin(\pi/n)}$ and then taking $n\to\infty$. Don't have to say this is not the recommended method $\endgroup$ – Winther Apr 22 '18 at 19:49

Note that $$\left|\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}-1 \right|\leq \left| \int_\limits{0}^{1-\delta}\frac{dx}{x^{n}+1} -(1-\delta)\right|+ \left| \int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1}-2 \delta \right|+\left| \int_\limits{1+\delta}^{\infty}\frac{dx}{x^{n}+1}\right|+3 \delta$$

and $$0 \leq \int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1} \leq \int_\limits{1-\delta}^{1+\delta}dx$$


$$\lim_{n\to\infty}\int_0^1\frac{dx}{x^n+1}=\int_0^1\frac{dx}{0+1}$$ and $$\lim_{n\to\infty}\int_1^\infty\frac{dx}{x^n+1}=\int_1^\infty0\,dx.$$ In both cases one can appeal to the Monotone Convergence theorem.

  • $\begingroup$ I am sorry but I am not supposed to use the Monotone Convergence theorem. $\endgroup$ – Pedro Gomes Apr 22 '18 at 15:01

Instead of working with the limit in the middle integral, treat it like $n$ is large, then approximate the integral (bound the integrand from above by $1$). You should end up with a direct $\delta$ dependence. If you let $\delta $ be arbitrary, what does that tell you the middle integral must be?


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