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I want to calculate $\int\limits_{0}^{\infty} \dfrac{\log(x)}{1+x^2}$. To do this, I consider $\int\limits_{\Gamma} \dfrac{\log^2(z)}{1+z^2}$ where $\Gamma$ is the keyhole contour. Basically all I need to do is to calculate $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big)$ and $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. For the first case we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=i} $ and $\log(i) = i\pi/2$. I'm struggling with the second case, namely with $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big)$. Again, we have $\text{Res}\Big(\dfrac{\log^2(z)}{1+z^2}, -i\Big) = \dfrac{\log^2(z)}{2z}\Big|_{z=-i} $ but what is $\log(-i)$? It seems to me that since we're working with the principal branch of logarithm we should have $\log(-i) = \log(|-i|) + i \arg(-i) = 3 \pi i/2$ since for the principal branch of logarithm one has $0 < \arg(z) < 2 \pi$. The "residue calculator" says that it should be $-i\pi/2$ and not $3i\pi/2$. Why?

iou
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  • See https://math.stackexchange.com/questions/76989/definite-integral-using-the-method-of-residues – Robert Z Dec 26 '20 at 18:15
  • Using real analysis I got $$\int \frac{\log (x)}{x^2+1} , dx=\log (x) \arctan x+\frac{1}{2} i (\text{Li}_2(-i x)-\text{Li}_2(i x))+C$$ Taking limits at $x\to 0$ and $x\to\infty$ result is zero. – Raffaele Dec 26 '20 at 19:10
  • You're absolutely right, it is $3\pi/2$. Indeed, if you calculated $\int_0^\infty\frac{dx}{1+x^2}$ (famously $\pi/2$) from $\oint_\Gamma\frac{\log zdz}{1+z^2}$, it would be the same story:$$-2\pi i\int_0^\infty\frac{dx}{1+x^2}=2\pi i\left(\frac{\pi i/2}{2i}+\frac{3\pi i/2}{-2i}\right)\implies\int_0^\infty\frac{dx}{1+x^2}=\frac{\pi}{2}.$$ – J.G. Dec 26 '20 at 19:48

2 Answers2

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I think the principal branch of logarithm is defined on the plane with the branch cut along the negative real axis (including zero), so the arguments lie in $(-\pi,\pi)$, directed from the positive real axis, hence $\log(-i) = -i\pi/2$

Alex Ortiz
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Your analysis is correct. If the branch cut is taken along the real axis, then the complex logarithm is defined by

$$\log(z)=\text{Log}(|z|)+i\arg(z)$$

where $\text{Log}$ is the logarithm function for real analysis and

$$0\le \arg(z)<2\pi$$

Hence, $\log(i)=i\pi/2$ and $\log(-i)=i3\pi/2$.

The residue calculator uses the principal branch of the logarithm, which is defined such which $-\pi<\arg(z)\le \pi$.

J.G.
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Mark Viola
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