I need help proving the following limit: $$\lim_{n \to \infty} \int_0^{\infty} \frac{1}{1+x^n}~dx = 1$$

In WolframAlpha I was playing around with the values of the sequence defined by the integral and noticed that the values seem to get arbitrarily close to 1. I guess the difficulty is finding a closed expression for the value of the definite integral.

  • $\begingroup$ Try using Dominated Convergence theorem. $\endgroup$ Jul 3, 2018 at 0:55
  • $\begingroup$ If $n$ is a positive even integer you should be able to evaluate the integral using complex analysis. $\endgroup$ Jul 3, 2018 at 1:03
  • $\begingroup$ A plot of the integrand with a large $n$ shows you that the function tends to a square. $\endgroup$
    – user65203
    Jul 3, 2018 at 7:47

7 Answers 7


Break it into two integrals, on $[0,1]$ and on $[1,\infty)$. $$\int_1^\infty\frac{dx}{1+x^n}<\int_1^\infty\frac{dx}{x^n}=\frac1{n-1}$$ so $$\int_1^\infty\frac{dx}{1+x^n}\to0.$$ Also $$1-\int_0^1\frac{dx}{1+x^n}=\int_0^1\frac{x^n}{1+x^n}\,dx <\int_0^1 x^n\,dx=\frac1{n+1}\to0$$ so $$\int_0^1\frac{dx}{1+x^n}\to1.$$

  • $\begingroup$ Quite good :). I'm just so used to use the dominated convergence theorem that I didn't try doing with standard techniques. $\endgroup$ Jul 3, 2018 at 1:05

Split the integral in two parts, one in $[0,1]$ and the second one in $[1,+\infty]$. Using the Dominated convergence theorem you should conclude that the first integral converges to $1$ and the second one to $0$. If you need more help let me know.



The integral

$$ \mathcal{I}= \int_0^\infty \frac{1}{1+x^n} dx,$$

can be equivalently expressed as

$$ \mathcal{I} = \frac{1}{n} \int^1_0 t^{\left(1-\frac{1}{n}\right) - 1} \left(1-t\right)^{\left(\frac{1}{n}\right)-1} dt = \frac{1}{n} B \left(1-\frac{1}{n},\frac{1}{n} \right),$$

where $B(x,y)$ is the Beta function. You can then make use of the identity

$$ B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{\Gamma\left( 1 - \frac{1}{n}\right) \Gamma \left(\frac{1}{n} \right)}{\Gamma(1)}, $$

where $\Gamma$ denotes the Gamma function and $\Gamma(1) = 1$. It can be shown that

$$ \Gamma\left( 1 - \frac{1}{n}\right) \Gamma \left(\frac{1}{n} \right) = \frac{\pi}{ \sin(\pi/n)}. $$

Hence, the integral takes the form

$$ \mathcal{I}= \int_0^\infty \frac{1}{1+x^n} dx = \left( \frac{\pi}{n} \right) \frac{1}{\sin(\pi/n)},$$

where the limit follows immediately.

The integral is obtained following the substitution

$$ t = \frac{1}{1+x^n}, $$

and making use of the fact

$$ dx = -\frac{1}{n} \left(\frac{1}{t(1-t)} \right) \left( \frac{1-t}{t}\right)^{1/n} dt.$$


For $x \le 1$ we have $x^{n+1} \le x^n$ so $\frac{1}{1+x^{n}} \le \frac1{1+x^{n+1}}$ so using the Lebesgue Monotone Convergence theorem we get

$$\lim_{n\to\infty}\int_{[0,1]}\frac{dx}{1+x^n} = \int_{[0,1]}\left(\lim_{n\to\infty}\frac{1}{1+x^n}\right)\,dx = \int_{[0,1]} dx = 1$$

For $x \ge 1$ and $n \ge 2$ we have $\frac{1}{1+x^n} \le \frac{1}{x^n} \le \frac1{x^2}$ which is integrable on $[1, +\infty)$ so using the Lebesgue Dominated Convergence Theorem we get

$$\lim_{n\to\infty}\int_{[1, +\infty)}\frac{dx}{1+x^n} = \int_{[1 ,+\infty)}\left(\lim_{n\to\infty}\frac{1}{1+x^n}\right)\,dx = \int_{[1, +\infty)} 0 \,dx = 0$$


$$\lim_{n\to\infty}\int_{[0, +\infty)}\frac{dx}{1+x^n} =1$$


The integral has the value

$$\frac{\pi \csc \left(\frac{\pi }{n}\right)}{n}$$

and the limit for $n \to \infty$ goes to $1$, where you can use l'Hospital's rule.

  • 3
    $\begingroup$ Well how'd you get that value? $\endgroup$
    – Wesley
    Jul 3, 2018 at 1:01
  • $\begingroup$ Using the beta function. @MilesDavis $\endgroup$ Jul 3, 2018 at 1:02
  • $\begingroup$ I think you must use some tricky substitution. $\endgroup$ Jul 3, 2018 at 1:02
  • $\begingroup$ @MilesDavis One way to evaluate the integral is to use contour integration with a wedge (pie-shaped) contour. A second methodology that uses real analysis only begins with the substitution $x\mapsto x^{1/n}$, recognizing of the resulting Beta function, using the relationship between Beta and Gamma functions, and exploiting the reflection property of the Gamma function. $\endgroup$
    – Mark Viola
    Jul 3, 2018 at 2:06
  • $\begingroup$ @MilesDavis Alternatively $$\int_0^1 \frac{1+x^{n-2}}{1+x^n}\,dx=\sum_{k=0}^\infty (-1)^k \left(\frac{1}{nk+1}+\frac{1}{nk+n-1}\right)$$which approaches $1$ as $n\to \infty$ $\endgroup$
    – Mark Viola
    Jul 3, 2018 at 2:27

Note that we can write for $n>1$

$$\begin{align} \int_0^\infty \frac{1}{1+x^n}\,dx&=\int_0^1 \frac{1+x^{n-2}}{1+x^n}\,dx\\\\ &=1+\int_0^1 \frac{x^{n-2}-x^n}{1+x^n}\,dx \end{align}$$


$$\left|\int_0^1 \frac{x^{n-2}-x^n}{1+x^n}\,dx\right|\le \frac{1}{n-1}-\frac{1}{n+1}$$

  • $\begingroup$ Would the anonymous down voter care to comment? $\endgroup$
    – Mark Viola
    Jul 14, 2018 at 18:11
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. If this answer is not useful, then I am happy to delete it. Would you please let me know? $\endgroup$
    – Mark Viola
    Sep 27, 2018 at 18:08
  • $\begingroup$ Hi Wesley. Would you please let me know how I can improve my answer? I really want to give you the best answer I can. If this was not useful, I am happy to delete it. Looking forward to your reply. Thank you in advance. $\endgroup$
    – Mark Viola
    Nov 17, 2018 at 18:34
  • $\begingroup$ And feel free to up vote any answer you found was useful - as you see fit of course. ;-) $\endgroup$
    – Mark Viola
    Nov 17, 2018 at 18:35

Consider the contour $[r,R], [R,Re^{\frac {2\pi}{n}i}], [Re^{\frac {2\pi}{n}i},re^{\frac {2\pi}{n}i}],[Re^{\frac {2\pi}{n}i}, r]$

$\displaystyle \int r^R f(x)\ dx + \int_0^{e^{\frac{2\pi}{n}i}} f(Re^{it})Rie^{it}\ dt +\int{R}^r f(e^{\frac{2\pi}{n}i}x)e^{\frac{2\pi}{n}}\ dx + \int_{e^{\frac{2\pi}{n}i}}^0 f(re^{it})rie^{it}\ dt = \oint_\gamma f(z)\ dz$

$\displaystyle\lim_\limits{R\to\infty}\int_0^{e^{\frac{2\pi}{n}i}} f(Re^{it})Rie^{it}\ dt = 0\\ \lim_\limits{r\to0}\int_{e^{\frac{2\pi}{n}i}}^0 f(re^{it})rie^{it}\ dt = 0$

$\displaystyle I - e^{\frac{2\pi}{n}i}I = \oint_\gamma f(z)\ dz$

There is one pole inside the contour.

$\oint_\gamma f(z)\ dz = 2\pi i(\frac {-e^{\frac{\pi}{n}i}}{n})\\ I = \frac {2 i }{n(e^{\frac{\pi}{n}i}- e^{-\frac{\pi}{n}i})} = \pi\frac {\csc \frac {\pi}{n}}{n}$


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.