6
$\begingroup$

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

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

$\endgroup$
  • 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
1
$\begingroup$

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 $$\psi=tan^{-1}[a^\frac n2]$$ $$Lt_{n\rightarrow\infty}\ \ \ [\psi]=\pi/2$$

$$Lt_{n\rightarrow\infty}[I_n]=\int_{\pi/4}^{\pi/2}\dfrac{2cos\theta}{sin\theta}.d\theta$$.

So,it can be done!

$\endgroup$
8
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.