$\lim_{n \to \infty} \displaystyle \int_1^a \dfrac{n}{1+x^n}dx$, where parameter $a>1$.

Some trick is probably needed here, but i don't see it.

  • 1
    $\begingroup$ Do you know contour integration? $\endgroup$ – Julien Clancy Mar 24 '13 at 15:40
  • $\begingroup$ Not really. But i will take a look! $\endgroup$ – Karlis Olte Mar 24 '13 at 15:44
  • $\begingroup$ It looks like the limit is $\log 2$, regardless of $a$. $\endgroup$ – Pedro Tamaroff Mar 24 '13 at 15:50

Hint: Put $$x=tan^{2/n}\theta$$ $$ dx=\dfrac2ntan^{\frac2n-1}\theta \times sec^2\theta . d\theta$$

Now => $$\boxed{I_n=\int_\phi^\psi 2tan^{\frac2n-1}\theta . d\theta}$$ $\phi,\psi$ are the limits accordingly. $$\phi=tan^{-1}1=\pi/4$$ and $$\psi=tan^{-1}[a^\frac n2]$$ $$Lt_{n\rightarrow\infty}\ \ \ [\psi]=\pi/2$$


So,it can be done!


$$\begin{align*}\int_1^a \frac{n}{1+x^n}\,dx &=\int_{1}^a\frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^a\frac{n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^a\frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}}-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{a^{1-n}}{n-1}-\frac{a^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{align*}$$

$$\text{Since}\; \lim_{n\to\infty}\frac{n}{kn-1}=\frac{1}{k}:$$

$$\lim_{n\to\infty}\int_1^a \frac{n}{1+x^n}dx=1-\frac{1}{2}+\frac{1}{3}-\cdots =\ln 2$$


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