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The question is calculate the value of the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ These are the steps that I followed,

Let $x$ be a complex number. So, the poles of the function $f(x)=\frac{1}{1+x^4}$ occur when $x$ is equal to the roots of the equation $1+x^4=0$, i.e $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$. They are all poles of degree 4.

Now Cauchy's theorem says that, $$\frac{1}{2\pi i}\int_C dx\ f(x)=\sum_i\text{Res}(f,x_i)$$ where $x_i$ are the poles of $f$ that lies within $C$. I am pretty sure that my poles lie within $-\infty$ and $\infty$. So, I calculated the residues of $\frac{1}{1+x^4}$ at $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$ and they are equal to $-\frac14e^{i\pi/4},-\frac14e^{3i\pi/4},-\frac14e^{5i\pi/4},-\frac14e^{7i\pi/4}$. Wolfram Alpha confirms my calculations.

The sum of residues is $$\sum_i\text{Res}=-\frac14e^{i\pi/4}-\frac14e^{3i\pi/4}-\frac14e^{5i\pi/4}-\frac14e^{7i\pi/4}=0$$ and therefore the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ must be equal to zero. However, Wolfram alpha says it is not zero but equal to$\frac{\pi}{\sqrt 2}$. Where am I making a mistake?

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In applying the residue theorem, we analyze the integral $I$ given by

$$\begin{align} I&=\oint_C \frac{1}{1+z^4}\,dz\\\\ &=\int_{-R}^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{1}{1+(Re^{i\phi})^4}\,iRe^{i\phi}\,d\phi\\\\ &=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right) \end{align}$$

where $R>1$ is assumed.


Note that the only residues implicated in the residue theorem are those enclosed by $C$. Here, $C$ is comprised of (i) the line segment from $-R$ to $R$ and (ii) the semicircular arc centered at the origin with radius $R$ and residing in the upper-half plane. Hence, the only resides are the ones at $z=e^{i\pi/4}$ and $z=e^{i3\pi/4}$.


Taking $R\to \infty$, the integration over the semi-circular contour vanishes and we find that

$$\begin{align} \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx&=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right)\\\\ &=2\pi i \left(-\frac{e^{i\pi/4}}{4}-\frac{e^{i3\pi/4}}{4}\right)\\\\ & =\frac{\pi}{\sqrt 2} \end{align}$$

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  • $\begingroup$ Like the other answer, why did you only used the upper half semicircle? Why did you not ignore the first term in the second step and evaluate over the whole circle like, $$\int_0^{2\pi} d\phi\ \frac{iRe^{i\phi}}{1+(Re^{i\phi})^4}$$ $\endgroup$ – sigsegv Apr 23 '17 at 17:13
  • $\begingroup$ @Ayatana We want to evaluate the integral $\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx$. How does integrating over the circle do that? Well, it doesn't; and that integral is $0$. So instead, we evaluate a contour integral, the one described herein, that has the real line integral as part of the contour. We could have taken the semi-circle in the lower-half plane, but since the orientation would be clockwise, we need a $-2\pi i$ factor o multiply the sum of the residues in the lower-half plane. The result will be the same, of course. ;-)) $\endgroup$ – Mark Viola Apr 23 '17 at 17:15
  • $\begingroup$ Is it necessary for the contour to be semicircle? Will a contour made of a straight line along real axis and an arbitrary line joing the ends work? $\endgroup$ – sigsegv Apr 24 '17 at 4:30
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    $\begingroup$ @Ayatana The reside theorem applies to any closed rectifiable curve. But we are free to choose the closed contour. And by choosing a semicircle the radius of which approaches infinity, we can show quite easily that its contribution approaches $0$ as $R\to \infty$. $\endgroup$ – Mark Viola Apr 24 '17 at 4:34
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You shouldn't need Wolfram Alpha to tell you the answer isn't zero. The integrand is positive, so the integral is positive, and certainly nonzero.

You have added up the residues at all the poles. However using the usual semicircle method gives you that the integral is $2\pi i$ times the sum of the residues of the poles in the upper half-plane.

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  • $\begingroup$ Why should I not add the residues of the poles below the lower half? What semi circle are you talking about? The limits of the integration are from $-\infty$ to $\infty$ for both real and imaginary parts of $x$. $\endgroup$ – sigsegv Apr 23 '17 at 16:29

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