$$\int_{0}^{\infty} \ln{(1+a x)}{x^{-b-1}} dx$$
I defined two branch cuts along the real axis: $[-\infty ,-\frac{1}{a}]$ & $[0,\infty]$ with the following contour: contour
I defined the $arg{(z)} =0$ above the positive branch cut and $arg(z)=2\pi$ below the positive branch cut. Similarly $arg(z)=\pi$ above the negative branch cut and $arg(z)=-\pi$ below the negative branch cut.

Using the triangle inequality for integrals it can easily (but with care) be shown that the integrals along all parts of the circles tend to $0$ as $R \Rightarrow \infty$ (radius of the outer circle) and $\epsilon \Rightarrow 0$ (radius of the smaller circles around the singularities).
By doing this we get the necessary bounds for a and b: $a>0$ and $0<b<1$

The integrals along the positive branch cut work out nicely and result in: $$(1-e^{-2 b \pi i}) \int_{0}^{\infty} \ln{(1+a x)}{x^{-b-1}} dx$$
When I work out the integrals along the I end up with the following for the integral along the top side of the negative branch cut: $$e^{-b \pi i} \int_{\infty}^{\frac{1}{a}} \ln{(1-a x)}{x^{-b-1}} dx$$
After factoring out $-1$ from the inside of the ln you are left with: $$e^{-b \pi i} \int_{\infty}^{\frac{1}{a}} (\ln{(a x-1)}-i \pi){x^{-b-1}} dx$$
Applying the same procedure to the integral along the bottom side of the negative branch cut yields: $$e^{b \pi i} \int_{\frac{1}{a}}^{\infty} (\ln{(a x-1)}+i \pi){x^{-b-1}} dx$$
This is troublesome because the $e^{b \pi i}$ stops me from combining the two integrals along the negative branch cut in order to cancel out the integrals involving the $\ln$
In the following article the author magically has $e^{-b \pi i}$ in front of the integral which allows him to cancel out the parts including the $\ln$ which simplifies it enormously.

Can someone please explain to me what I am doing wrong with the integrals along the negative branch cut?

Any help is greatly appreciated!


2 Answers 2


On the negative branch cut, $\arg z$ stays continuous (and is equal to $\pi$; it is $\arg(1+az)$ that gets discontinuous). So your last two integrals should get $e^{-b\pi i}$ both.

  • $\begingroup$ Thank you for your quick response!! $\endgroup$
    – M.Bakkers
    May 3, 2019 at 18:34
  • $\begingroup$ Could you please explain to me why I can simply set the argument of z along the negative branch cut to be $\pi$ because you could have chosen $-\pi$ as well. What is the underlying reason that I have to set the arg to be $\pi$ rather than $-\pi$? $\endgroup$
    – M.Bakkers
    May 3, 2019 at 19:15
  • 1
    $\begingroup$ You have chosen the branch of $\arg z$ which is $0$ above the positive cut and is $2\pi$ below it. For negative real $z$ this branch meets $\arg z=\pi$. Choosing $-\pi$ instead means choosing a different branch (the one with $\arg z=-2\pi$ above the positive cut and $\arg z=0$ below it). $\endgroup$
    – metamorphy
    May 4, 2019 at 5:56

I also have a hard time managing double branch cuts, so to overcome this I often use Differentiation Under the Integral Sign (DUIS) to avoid it. First, let’s rewrite $log(1+ax)$ as $\int_{0}^{a}\frac{x}{1+yx}dy$: $$I=\int_{0}^{\infty}log(1+ax)x^{-b-1}dx=\int_{0}^{a}\int_{0}^{\infty}\frac{x^{-b}}{1+yx}dxdy\overbrace{=}^{yx\rightarrow z}\int_{0}^{a}y^{b-1}dy\int_{0}^{\infty}\frac{z^{-b}}{1+z}dz$$ $$I=\left[\frac{y^{b}}{b}\right]_0^a \mathfrak{B}(1-b,b)=\frac{\pi a^{b}}{b} csc(\pi b)$$

$\textbf{Addendum:}$ In this case it was possible to use the Beta Function instead of Complex Analysis, however if the function obtained wasn't as nice as this one, I would compute the residues of the integral obtained after applying DUIS.


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