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.
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 $$\int_0^1\frac{dx}{1+x^n}\to1.$$
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.
HINT
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
$$\lim_{n\to\infty}\int_{[1, +\infty)}\frac{dx}{1+x^n} = \int_{[1 ,+\infty)}\left(\lim_{n\to\infty}\frac{1}{1+x^n}\right)\,dx = \int_{[1, +\infty)} 0 \,dx = 0$$
Therefore
$$\lim_{n\to\infty}\int_{[0, +\infty)}\frac{dx}{1+x^n} =1$$
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.
Note that we can write for $n>1$
$$\begin{align} \int_0^\infty \frac{1}{1+x^n}\,dx&=\int_0^1 \frac{1+x^{n-2}}{1+x^n}\,dx\\\\ &=1+\int_0^1 \frac{x^{n-2}-x^n}{1+x^n}\,dx \end{align}$$
and
$$\left|\int_0^1 \frac{x^{n-2}-x^n}{1+x^n}\,dx\right|\le \frac{1}{n-1}-\frac{1}{n+1}$$
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.
$\oint_\gamma f(z)\ dz = 2\pi i(\frac {-e^{\frac{\pi}{n}i}}{n})\\ I = \frac {2 i }{n(e^{\frac{\pi}{n}i}- e^{-\frac{\pi}{n}i})} = \pi\frac {\csc \frac {\pi}{n}}{n}$
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