# Finding $lim_{n to infty} int_1^a frac{n}{1+x^n} , dx$

Mathematics Asked by Karlis Olte on January 3, 2021

I’m trying to find $$lim_{n to infty} int_1^a dfrac{n}{1+x^n} , dx ,$$ where parameter $$a>1$$.

The limit appears to be $$ln 2$$ regardless of the value of $$a$$.

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

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

Now => $$boxed{I_n=int_phi^psi 2tan^{frac2n-1}theta . dtheta}$$ $phi,psi$ are the limits accordingly. $$phi=tan^{-1}1=pi/4$$ and $$psi=tan^{-1}[a^frac n2]$$ $$Lt_{nrightarrowinfty} [psi]=pi/2$$

$$Lt_{nrightarrowinfty}[I_n]=int_{pi/4}^{pi/2}dfrac{2costheta}{sintheta}.dtheta$$.

### So,it can be done!

Correct answer by ABC on January 3, 2021

EDIT: I tried to make my answer more rigorous.

Let $$u = x^{n}$$.

Then we have

$$lim_{n to infty} int_1^a frac{n}{1+x^n} , mathrm dx =lim_{n to infty} int_{1}^{a^{n}} frac{u^{1/n-1}}{1+u} , mathrm du = lim_{n to infty} int_{1}^{infty} boldsymbol{1}_{[1,a^{n}]}(u) , frac{u^{1/n-1}}{1+u} , mathrm du.$$

For $$n ge 2$$, the integrand is dominated by the integrable function $$frac{u^{-1/2}}{1+u}$$.

So by appealing to the the dominated convergence theorem, we can conclude that

$$lim_{n to infty} int_1^a frac{n}{1+x^n} , mathrm dx = int_{1}^{infty} frac{u^{-1}}{1+u} , mathrm du = int_{1}^{infty} left(frac{1}{u} - frac{1}{1+u} right) mathrm du = ln 2.$$

Answered by Random Variable on January 3, 2021

begin{align*}int_1^a frac{n}{1+x^n},dx &=int_{1}^afrac{n}{x^n}left(frac{1 }{1+(1/x)^n}right)dx\&=int_1^afrac{n}{x^n}left(1-frac{1}{x^n}+frac{1}{x^{2n}}-cdotsright)dx\&=nint_1^afrac{1}{x^n}-frac{1}{x^{2n}}+frac{1}{x^{3n}}-cdots,dx\&=nleft[frac{1}{n-1}-frac{1}{2n-1}+cdotsright]-underbrace{nleft[frac{a^{1-n}}{n-1}-frac{a^{1-2n}}{2n-1}+cdotsright]}_{to,0}end{align*}

$$text{Since}; lim_{ntoinfty}frac{n}{kn-1}=frac{1}{k}:$$

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

Answered by L. F. on January 3, 2021