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  • Problem 5.16

    [5.16] Compute\(\int_{-\infty}^{\infty}\frac{dx}{x^{4}+1}\)

    Let \(\sigma_R\) be the semicircle, oriented counter-clockwise, formed by the segment \([-R,R]\) on the real axis and the circular arc, \(\gamma_R\), of radius \(R\) in the upper half of the 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 the first quadrant. 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 the second quadrant.
    Now consider the complex integral: \[\begin{aligned} \int_{\sigma_R}\frac{dz}{z^{4}+1} &= \int_{\gamma_1}\frac{dz}{z^{4}+1} + \int_{\gamma_2}\frac{dz}{z^{4}+1}\end{aligned}\] Now first look at the integral over \(\gamma_1\). \[\begin{aligned} \int_{\gamma_1}\frac{dz}{z^{4}+1} &= \int_{\gamma_1}\frac{dz}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}\\ &= \int_{\gamma_1}\frac{\frac{dz}{(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))}\end{aligned}\] Since the function in the numerator is holomorphic in \(\gamma_1\), we can use Cauchy’s Integral formula. So we have, \[\begin{aligned} \int_{\gamma_1}\frac{\frac{dz}{(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))} &= \frac{2\pi i}{(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}+\frac{\sqrt2}{2}-i\frac{\sqrt2}{2})(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}+\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})}\\ &= \frac{2\pi i}{(\sqrt2)(\sqrt2+i\sqrt2)(i\sqrt2)}\\ &= \frac{2\pi i}{2i(\sqrt2 +i\sqrt2)}\\ &= \frac{2\pi i}{2i\sqrt2-2\sqrt2}\\ &=\frac{\pi i}{-\sqrt2+i\sqrt2}\\\end{aligned}\] And now for the integral over \(\gamma_2\). \[\begin{aligned} \int_{\gamma_1}\frac{dz}{z^{4}+1} &= \int_{\gamma_1}\frac{dz}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}\\ &= \int_{\gamma_1}\frac{\frac{dz}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}}{(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))}\end{aligned}\] We may use Cauchy’s Integral Formula once again because the function in the numerator is holomorphic in \(\gamma_2\). \[\begin{aligned} \int_{\gamma_1}\frac{\frac{dz}{(z-(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))(z-(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))(z-(\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}))}}{(z-(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}))} &= \frac{2\pi i}{(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2})(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}+\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})(-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}-\frac{\sqrt2}{2}+i\frac{\sqrt2}{2})}\\ &= \frac{2\pi i}{(-\sqrt2)(-\sqrt2+i\sqrt2)(i\sqrt2)}\\ &= \frac{2\pi i}{-2i(-\sqrt2 +i\sqrt2)}\\ &= \frac{2\pi i}{2i\sqrt2+2\sqrt2}\\ &=\frac{\pi i}{\sqrt2+i\sqrt2}\\\end{aligned}\] So we have, \[\begin{aligned} \int_{\sigma_R}\frac{dz}{z^{4}+1} &= \int_{\gamma_1}\frac{dz}{z^{4}+1} + \int_{\gamma_2}\frac{dz}{z^{4}+1}\\ &= \frac{\pi i}{-\sqrt2+i\sqrt2} + \frac{\pi i}{\sqrt2+i\sqrt2}\\ &= \frac{\pi i(\sqrt2 + i\sqrt2)+ \pi i(-\sqrt2 + i\sqrt2)}{(-\sqrt2 + i\sqrt2)(\sqrt2 + i\sqrt2)}\\ &= \frac{\pi i(2i\sqrt2)}{-2-2}\\ &= \frac{-2\pi\sqrt2}{-4}\\ &= \frac{\pi\sqrt2}{2}\\\end{aligned}\] Now using Proposition 4.8(d) and the reverse triangle inequality, \[\begin{aligned} \left|\int_{\gamma_R}\frac{dz}{z^{4}+1}\right| &\leq max_{z\in\gamma_R}\left|\frac{1}{z^{4}+1}\right|\pi R\\ &\leq max_{z\in\gamma_R}\left(\frac{1}{|z|^{4}-1}\right) = \frac{\pi R}{R^{4}-1} \end{aligned}\] Which goes to zero as R approaches infinity. Thus, \[\begin{aligned} \frac{\pi\sqrt2}{2} &= \lim_{R\rightarrow\infty}\int_{\sigma_R}\frac{dz}{z^{4}+1}\\ &= \lim_{R\rightarrow\infty}\int_{[-R,R]}\frac{dz}{z^{4}+1}+\lim_{R\rightarrow\infty}\int_{\gamma_R}\frac{dz}{z^{4}+1}\\ &=\lim_{R\rightarrow\infty}\int_{[-R,R]}\frac{dz}{z^{4}+1}\\ &=\int_{-\infty}^{\infty}\frac{dx}{x^{4}+1}\end{aligned}\]

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