Let $$S_n = \int_{0}^{1} \frac{nx^{n-1}}{1+x}dx$$ for $n >0$. Then as $n \to \infty$ , the sequence $(S_n)_{n>0}$ tends to

  1. $0$

  2. $1/2$

  3. $1$

  4. $+\infty$

$$S_n = \int_{0}^{1} \frac{nx^{n-1}}{1+x}dx$$ put $x= \tan^{2}t$

$dx = 2 \tan t \sec^{2} t dt$

so, $$S_n = \int_{0}^{\pi/4} \frac{n \tan^{2n-2}t}{1+\tan^{2} t} 2 \tan t \sec^{2}t dt$$ $\int_{0}^{\pi/4} 2n \tan^{2n-1}t dt$

$2n\int_{0}^{\pi/4} \tan^{2n-1}t dt$

I don't know how to proceed further to find the limit. Is there any other method to find the limit?

  • 1
    $\begingroup$ If you just want to proceed by elimination, $$\frac{1}{2} \int_0^1 nx^{n-1}dx \leq S_n \leq \int_0^1 nx^{n-1}dx$$ immediately removes 1. and 4. $\endgroup$ – Clement C. Dec 24 '18 at 7:41

Integrating by parts we get $S_n=x^{n} \frac 1 {1+x} |_0^{1}+\int_0^{1} x^{n} \frac 1 {(1+x)^{2}}$. Second term tends to $0$ so the answer is $\frac 1 2$. [Second term is $\leq \int_0^{1}x^{n}dx =\frac 1 {n+1}$].

  • $\begingroup$ Thanks! It is really easy than what i was trying to do. $\endgroup$ – Mathsaddict Dec 24 '18 at 6:46

A third answer, just to see more techniques.

From the change of variables $u=x^n$ (so that $du = n x^{n-1}dx$), we have $$ S_n = \int_0^1 \frac{du}{1+u^{1/n}} $$ and then, by your favorite convergence theorem for integrals (in my case, the Dominated Convergence Theorem), we have $$ \lim_{n\to\infty} S_n = \int_0^1 du\,\lim_{n\to\infty}\frac{1}{1+u^{1/n}} = \int_0^1 du\cdot \frac{1}{2} = \boxed{\frac{1}{2}} $$

  • $\begingroup$ I have a little confusion about taking the limit under integral sign. Can we always move the limit sign under integration or there is some condition to do this? $\endgroup$ – Mathsaddict Dec 24 '18 at 8:00
  • $\begingroup$ There are conditions. Here, it's legitimate to do it because of the Dominated Convergence Theorem (it's the result I choose to invoke to justify this step here). @Mathsaddict $\endgroup$ – Clement C. Dec 24 '18 at 8:02

Since you already received a good answer from Kavi Rama Murthy, I add a few thinks for your curiosity.

The antiderivative $$I_n = \int\frac{n\,x^{n-1}}{1+x}\,dx$$ can be computed using special functions that you will learn sooner or later. From it, the integral $$S_n = \int_{0}^{1} \frac{n\,x^{n-1}}{1+x}\,dx=\frac{1}{2} n \left(\psi \left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)\right)$$ where appears the digamma function (which is "similar" to harmonic numbers). Using their asymptotics, we have $$S_n=\frac{1}{2}+\frac{1}{4 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached as well as an approximation.

For example, using $n=10$, the exact result would be $\approx 0.524877$ while the aboce formula gives $\frac{4199}{8000}\approx 0.524875$.


In context:


$\displaystyle{\int_{0}^{1}}\dfrac{nx^{n-1}}{1+1}dx \lt I_n\lt$

$\displaystyle{\int_{0}^{1/2}}\dfrac{nx^{n-1}}{1}dx + \displaystyle {\int_{1/2}^{1}}\dfrac{nx^{n-1}}{1+1/2}dx.$

$1/2< I_n <$

$(1/2)^n +(2/3)[1-(1/2)^n]=(1/3)(1/2)^n+2/3.$

Taking the limit $n \rightarrow \infty:$

$1/2 \le \lim_{n \rightarrow \infty}I_n \le 2/3.$

Assuming one of the answers is correct, it is answer 2).


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.