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I’m trying to understand how to solve the following integral:

$$\int_0^{\infty}\frac{x^{1/4}}{x^2+1}dx$$

From my understanding I need to draw a Branch cut but I’m having troubles understanding where to do it in order to set up my contour to integrate using the residue theorem.

I understand how it works for a normal square root but here I’m a bit lost.

This is what I’ve tried:

The function has simple poles at $z=\pm i$ and this is the contour I tried:

$$\Gamma_1 = \big\{z=Re^{i\theta}, 0<\theta<2\pi, R>0\big\}$$

$$\Gamma_2= \big\{z=re^{i2\pi}, \epsilon<r<R\big\}$$

$$\Gamma_3 = \big\{z=\epsilon e^{i\theta}, 0<\theta<2\pi\big\}$$

$$\Gamma_4 = \big\{z=re^{i0}, \epsilon<r<R\big\}$$

So a “keyhole” contour as my professor says.

Due to the estimation Lemma and the fact that the function is regular, we ignore the contributions of $\Gamma_{1,3}$

On $\Gamma_2$ we end up with $-iI$ where $I$ is the integral.

On $\Gamma_1$ we end up with $I$ and using the Residue theorem:

$$I(1-i) = 2\pi \cos(\pi/8)$$.

This seems to not be the actual answer and I don’t know where I went wrong.

Many thanks!

bsaoptima
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  • I'm pretty sure taking the branch cut along $[0,\infty)$ should work – Lorago Jan 11 '23 at 16:53
  • I tried it but I got a wrong result. – bsaoptima Jan 11 '23 at 16:54
  • Could you include your attempt in the post so that we can see where it goes wrong? – Lorago Jan 11 '23 at 17:08
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    The analysis is the same as in the square root case. Above the cut $[0,\infty)$, $\arg(z)\to0$, while below the cut, $\arg(z)\to2\pi$. So in the first case, $z^{1/4}\to x^{1/4} e^{i\frac14\cdot0} = x^{1/4}$, while in the second, $z^{1/4}\to x^{1/4} e^{i\frac14\cdot2\pi} = ix^{1/4}$. (Had the integrand instead involved $x^{1/2}$, you'd end up with $x^{1/2}$ on one side of the cut and $-x^{1/2}$ on the other.) – user170231 Jan 11 '23 at 17:36
  • @user170231 this is exactly what I did but then I ended up with a different result than the one required. I’m writing my work as we speak. – bsaoptima Jan 11 '23 at 17:38
  • @Lorago i edited my post. – bsaoptima Jan 11 '23 at 17:42
  • Your residue sum seems to be wrong. I suspect you took the wrong arguments for the poles - remember, cutting along $[0,\infty)$ enforces $\arg(z)\in(0,2\pi)$, so $i=e^{i\frac\pi2}$ but $-i=e^{\color{red}{i\frac{3\pi}2}}$ – user170231 Jan 11 '23 at 17:45
  • @user170231 but isn’t $(-i)^{1/4}= \cos(3 \pi/8) +i\sin(3\pi/8)= \cos( \pi/8) -i\sin(\pi/8)$ – bsaoptima Jan 11 '23 at 17:49
  • Wait no I’m so dumb – bsaoptima Jan 11 '23 at 17:51
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    No, according to the restriction of the cut,$$(-i)^{1/4}=|-i|^{1/4}e^{i\frac14\cdot\frac{3\pi}2}=e^{i\frac{3\pi}8}=ie^{-i\frac\pi8}=\sin\left(\frac\pi8\right)+i\cos\left(\frac\pi8\right)$$ – user170231 Jan 11 '23 at 17:53
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    why not first substitute $x=y^4$ and then you get an easy residue integral of a rational function with $4$ simple poles in the upper plane at roots of unity for which the residues are not that hard to compute – Conrad Jan 11 '23 at 17:54
  • In THIS ANSWER, I used contour integration to evaluate the integral $\int_{0}^\infty \frac{x^a}{1+x^2},dx$ for $|a|<1$ – Mark Viola Feb 04 '23 at 19:53

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