5.16
5.16
Compute \(\int_{-\infty}^{\infty} \frac{dx}{x^4+1}\).
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Let \(\sigma_R\) be a semicircle oriented counter-clockwise, formed by the segment \([-R,R]\) on the Real-axis and the circular are, \(\gamma_R\), of the radius \(R\) in the positive Complex plane. Let \(\gamma_1\) be the quarter circle, oriented counter-clockwise, whose path is given by the Real and Imaginary axes and the portion of \(\gamma_R\) in Quadrant I. Let \(\gamma_2\) be the quarter circle, oriented counter-clockwise, whose boundary is the Real and Imaginary axes and the portion of \(gamma_R\) in Quadrant II.
Consider the Complex integral: \[\int_{\sigma_R}\frac{dz}{z^4+1}=\int_{\gamma_1}\frac{dz}{z^4+1}+\int_{\gamma_2}\frac{dz}{z^4+1}\] So the integral over \(\gamma_1\) is: \[\begin{aligned} \int_{\gamma_1}\frac{dz}{z^4+1}&=\int_{\gamma_1} \frac{dz}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))}\\ &=\int_{\gamma_1}\frac{\frac{dz}{(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(-\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{-2}}{2}))}}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))}\end{aligned}\] Because the numerator is a holomorphic function inside \(\gamma_1\), we can use Cauchy’s integral formula: \[\begin{aligned} \int{\gamma_1}\frac{\frac{dz}{(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(-\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{-2}}{2}))}}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))}&=\frac{2\pi}{(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}} +\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})\\ &(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\\ &(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\\ &=\frac{2\pi i}{(\sqrt{2})(\sqrt{2}+i\sqrt{2})(i\sqrt{2}}\\ &=\frac{2 \pi i}{2i(\sqrt{2}+i\sqrt{2})}\\ &=\frac{2 \pi i}{2i\sqrt{2}-2\sqrt{2}}\\ &=\frac{\pi i}{-\sqrt{2}+i\sqrt{2}}\end{aligned}\] And the integral over \(\gamma_2\) is: \[\begin{aligned} \int_{\gamma_2}\frac{dz}{z^4+1}&=\int_{\gamma_2}\frac{dz}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))}\\ &=\int{\gamma_2}\frac{\frac{dz}{(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(-\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{-2}}{2}))}}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))}\end{aligned}\] Again, we use Cauchy’s integral formula because inside \(\gamma_2\) is holomorphic: \[\begin{aligned} \int_{\gamma_2}\frac{\frac{dz}{(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))(z-(-\frac{-\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}))(z-(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{-2}}{2}))}}{(z-(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}))}&=\frac{2\pi}{(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}} +\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})\\ &(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\\ &(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\\ &=\frac{2\pi i}{(\sqrt{2})(\sqrt{2}+i\sqrt{2})(i\sqrt{2}}\\ &=\frac{2 \pi i}{2i(\sqrt{2}+i\sqrt{2})}\\ &=\frac{2 \pi i}{2i\sqrt{2}-2\sqrt{2}}\\ &=\frac{\pi i}{-\sqrt{2}+i\sqrt{2}}\end{aligned}\] So we have: \[\begin{aligned} \int_{\sigma_R}\frac{dz}{z^4+1}&=\int_{\gamma_1}\frac{dz}{z^4+1}+\int_{\gamma_2}\frac{dz}{z^4+1}\\ &=\frac{\pi i}{-\sqrt{2}+i\sqrt{2}}+\frac{\pi i}{\sqrt{2}+i\sqrt{2}}\\ &=\frac{\pi i(\sqrt{2}+i\sqrt{2})+\pi i(-\sqrt{2}+i\sqrt{2})}{(-\sqrt{2}+i\sqrt{2})(\sqrt{2}+i\sqrt{2})}\\ &=\frac{\pi i (2i\sqrt{2})}{-2-2}\\ &=\frac{=2\pi(\sqrt{2})}{-4}\\ &=\frac{\pi\sqrt{2}}{2}\end{aligned}\] Then by Proposition 4.8d and the reverse triangle inequality we have: \[\begin{aligned} \mid\int_{\gamma_R}\frac{dz}{z^4+1}\mid&\leq max_{z \in {\gamma_R}} \mid \frac{1}{z^4+1} \mid * \pi * R\\ &\leq max_{z \in {\gamma_R}} \frac{1}{\mid z\mid^4-1} = \frac{\pi R}{R^4-1}\end{aligned}\] And as \(R\) approaches infinity, this goes to zero. Therefore: \[\begin{aligned} \frac{\pi \sqrt{2}}{2}&=lim_{R \rightarrow \infty}\int_{\sigma_R}\frac{dz}{z^4+1}\\ &=lim_{R \rightarrow \infty}\int_{-R,R}\frac{dz}{z^4+1}+lim_{R \rightarrow \infty}\int_{\gamma_R}\frac{dz}{z^4+1}\\ &=lim_{R \rightarrow \infty}\int_{-R,R}\frac{dz}{z^4+1}+0\\ &=\int_{-\infty}^{\infty}\frac{dz}{z^4+1}\end{aligned}\]