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I'm supposed to evaluate:

$$ \int_0^{\infty} \frac{1}{x^4+1} dx $$

Consider $$ \oint \frac{1}{z^4+1} dz = \oint \frac{1}{(z - \frac{1-i}{\sqrt 2})(z + \frac{1-i}{\sqrt 2})(z - \frac{1+i}{\sqrt 2})(z + \frac{1+i}{\sqrt 2})} dz$$

The poles are at $z = \pm \frac{1-i}{\sqrt 2} $ and $z = \pm \frac{1+i}{\sqrt 2} $.

Consider this contour:

enter image description here

Take the semi-circle to infinity, the integral along the chord goes to zero, and integral becomes:

$$\int_{-\infty}^{\infty} \frac{1}{x^4 +1} dx $$

Residue at $z=\frac{i-1}{\sqrt 2}$ is $\frac{i}{2\sqrt 2(i-1)}$. Residue at $z=\frac{1+i}{\sqrt 2}$ is $\frac{-i}{2 \sqrt 2 (i+1)}$. Sum of residue is $\frac{-i}{2\sqrt 2}$.

$$\int_{-\infty}^{\infty} \frac{1}{x^4 +1} dx = \pi i \times \frac{-i}{2\sqrt 2} = \frac{\pi}{2\sqrt 2} $$

$$\int_{0}^{\infty} \frac{1}{x^4 +1} = \frac{\pi}{4\sqrt 2} $$.

Using, wolframalpha, the answer is twice of mine. Wolframalpha is never wrong.

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    $\begingroup$ I wonder why you don't integrate on the boundary of the first quadrant and for the path running along the positive imaginary axis notice $(iy)^4 = y^4$, so the two integrals along rays are the same... $\endgroup$
    – Eric Towers
    Jun 15, 2014 at 22:45

1 Answer 1

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You can't use the contour you tried to use, because the real integral you want to compute is not on the contour. Instead you should use a semicircle with the diameter on the real axis. Then the poles are inside the contour, so their residues give you the factor of $2 \pi i$ you need to get the right answer.

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    $\begingroup$ I've thought about that, but the loops enclosing the points are semicircles, not full circles. $\endgroup$
    – user44840
    Jun 15, 2014 at 22:40
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    $\begingroup$ Oh, I misunderstood what contour you used in the first place. You can't use that contour, because the real integral you want to compute is not on the contour. Instead your contour should simply be a semicircle with the diameter on the real axis. Then the residues are contained in the contour, so you have the factor of $2 \pi i$ as needed. In short, you made the problem harder than it is. $\endgroup$
    – Ian
    Jun 15, 2014 at 22:42
  • $\begingroup$ Brilliant, thank you. Yeah you're right, because along the horizontal line, $z = x + \frac{i}{\sqrt 2}$. $\endgroup$
    – user44840
    Jun 15, 2014 at 22:48
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    $\begingroup$ @Ian if you update your answer, (from your comment), I will upvote this post so it comes off of the unanswered queue. $\endgroup$
    – dustin
    Feb 25, 2015 at 23:45

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