Hard Integration

We can use contour integration to solve the integral, with a closed upper semicircular arc in the complex plane with its radius $R \to \infty$ as contour $\gamma$ .

$\begin{aligned} I & = \int_0^\infty \frac 1{x^4+1} dx & \small \blue{\text{As the integrand is even.}} \\ & = \frac 12 \int_{-\infty}^\infty \frac 1{x^4+1} dx & \small \blue{\text{Changing }x \text{ to complex variable }z} \\ & = \frac 12 \int_\gamma \frac 1{z^4+1} dz & \small \blue{\text{and integrate over contour }\gamma.} \\ & = \frac 12 \int_\gamma \frac {dz}{\blue{(z-e^{\frac \pi 4 i})(z-e^{\frac {3\pi 4}i})}(z+e^{\frac \pi 4 i})(z+e^{\frac {3\pi 4}i})} & \small \blue{\text{Only 2 of the 4 poles in contour }\gamma} \end{aligned}$

$\begin{aligned} \ \ & = \small \frac 1{2\sqrt 2} \left( \int_\gamma \frac {dz}{\blue{\left(z-\frac 1{\sqrt 2} -\frac i{\sqrt 2} \right)}\left(z+\frac 1{\sqrt 2} +\frac i{\sqrt 2} \right)\left(z-\frac 1{\sqrt 2} +\frac i{\sqrt 2} \right)} - \int_\gamma \frac {dz}{\blue{\left(z+\frac 1{\sqrt 2} -\frac i{\sqrt 2} \right)}\left(z+\frac 1{\sqrt 2} +\frac i{\sqrt 2} \right)\left(z-\frac 1{\sqrt 2} +\frac i{\sqrt 2} \right)} \right) \\ & = \frac {2\pi i}{2\sqrt 2}\left(\frac 1{2(i-1)} + \frac 1{2(1+i)}\right) \quad \quad \small \blue{\text{Applying Cauchy integral formula}} \\ & = \frac \pi{2\sqrt 2} \approx \boxed{1.1107} \end{aligned}$

For $n\in\mathbb N$ $I(n)=\int_0^{\infty} \frac{1}{1+x^n} dx =\frac{\pi}{n} \csc\left(\frac{\pi}{n}\right)$ and hence for $n=4$ we have $I(4)=\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right)\approx 1.1107$

Generalization For all $n>1$ we observe that $I(n)= \int_{0}^{\infty}\dfrac{\,dx}{1+x^n}$ is convergent. Now making u-substitution of the integrand we get $I(n) = -\int_{1}^{0} u\dfrac{(1+x^n)^2}{nx^{n-1}}\,du$ as $u=\,(1+x^n)^{-1}$ and reversing the limits we have then $\begin{aligned} I(n) & = \dfrac{1}{n} \int_{0}^{1}\dfrac{u}{u^2}\cdot\dfrac{u^{-\frac{1}{n}+1}}{(1-u)^{-\frac{1}{n} +1}}\,du \\&= \dfrac{1}{n}\int_{0}^{1}\dfrac{u^{-\frac{1}{n}}}{(1-u)^{-\frac{1}{n}+1}}\,du\end{aligned}$ giving us $\dfrac{B}{n}\left(1-\dfrac{1}{n} , \dfrac{1}{n}\right)=\dfrac{1}{n}\dfrac{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(\frac{1}{n}\right)}{\Gamma(1)}$ and hence by Euler reflection formula we have $I(n) =\dfrac{1}{n}\dfrac{\pi}{\sin \frac{\pi}{n}}$