Intereting Posts

How to build a function that gives recent years higher weight?
Is there a “deep line” topological space in analogue to the “long line” $\omega_1\times[0,1)$?
Generalized Euler phi function
First-order nonlinear ordinary differential equation
Show linear independence of a set of vectors close to an orthonormal basis
Tiling Posters on a Wall
If $\gcd(a,b)=1$, $\gcd(a,y)=1$ and $\gcd(b,x)=1$ then prove that $ax+by$ is prime to $ab$
How find all ordered pairs $(m,n)$ such $\frac{n^3-1}{mn-1}$ is an integer
Given $n\in \mathbb N$, is there a free module with a basis of size $m$, $\forall m\geq n$?
Limit of integral – part 2
Probability, conditional on a zero probability event
For natural numbers $a$ and $b$, show that $a \Bbb Z + b \Bbb Z = \gcd(a, b)\Bbb Z $
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Prove $f=x^p-a$ either irreducible or has a root. (arbitrary characteristic) (without using the field norm)
Another interesting integral related to the Omega constant

Many of us have seen the evaluation of the integral

$$\int^{\infty}_0 \frac{dx}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

It can be solved using contour integration or beta function .

- existence of sequence of polynomial
- Making a cube root function analytic on $\mathbb{C}\backslash $
- What is a geometric explanation of complex integration in plain English?
- roots of a polynomial inside a circle
- Limit point of sequence vs limit point of the set containing all point of the sequence
- erf(a+ib) error function separate into real and imaginary part

I thought of how to solve the integral

$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

It can be solved using real methods as follows

consider the following integral

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$

Differentiating with respect to $s$ we get

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1}\log(1+x) dx=\frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)} \left(\psi_0 (1-s)- \psi_0(p-s)\right)$$

at $s =0$ we get

$$\int^{\infty}_0 x^{-p}\frac{\log(1+x)}{1+x} dx=\frac{\pi}{\sin(\pi p)} \left(\psi_0 (1)- \psi_0(p)\right)$$

where i used the reflection formula .

**Statement of question**

How to solve the following integral using contour integration

$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

I thought we can use the following contour

So the function

$$F(z) = \frac{e^{-p \log(z)}\log(1+z)}{(1+z)} $$

is analytic in and on the contour by choosing the branch cut of $e^{-p \log(z)}$ as $0\leq \text{Arg}(z)<2\pi$ and the branch cut of $\log(1+z)$ as $0\leq \text{Arg}(z+1)<2\pi$ so the function $F(z)$ is analytic everywhere except at $z\geq -1$ . I am finding difficulty finding the integral on the branch point $z=-1$ it seems there is a contribution of the branch point and the pole .

Please don’t make any substitutions or simplifications for the integral. Feel free to use another contour if my choice was wrong .

- Conformal map from a lune to the unit disc in $\mathbb{C}$
- Convergence of infinite product $\prod_{n=2}^\infty (1- \frac 1n) $
- the product of diagonals of a regular polygon
- Laurent series for $1/(e^z-1)$
- Limit of argz and r
- $f(z)$ and $\overline{f(\overline{z})}$ simultaneously holomorphic
- Entire function with positive real part is constant (no Picard)
- Approximate spectral decomposition
- Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$
- Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$

To evaluate the integral using contour integration, consider the integral

$$\oint_C dz \frac{z^{-p} \log{(1+z)}}{1+z} $$

where $C$ is the following contour:

The magnitude of the integral about the large arc of radius $R$ behaves as $\frac{\log{R}}{R^p}$ as $R \to \infty$ and thus vanishes. Let the radius of the small circular arcs be $\epsilon$. The contour integral is then equal to, in this limit,

$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} – e^{-i \pi p}\int_{\infty}^{1+\epsilon} dx \frac{x^{-p} [\log{(x-1)}+i \pi]}{1-x} \\ – e^{-i \pi p}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p} [\log{(x-1)}-i \pi]}{1-x}+i \epsilon \int_{\pi}^{-\pi} d\phi \,e^{i \phi} \frac{(e^{i \pi}+\epsilon e^{i \phi})^{-p} \log{(\epsilon e^{i \phi})}}{\epsilon e^{i \phi}}$$

The first integral represents the integral about the branch cut along the positive real axis, which concerns the $z^{-p}$ term only. Note that the integral about the origin vanishes as $\epsilon \to 0$. The second and third integrals represent the integrals along each side of the branch cut on the negative axis. Note that, along this branch cut, the argument of $z^{-p}$ is $-\pi p$ on either side of the branch cut, as the branch cut there concerns the log term only. The fourth integral is the integral about the branch point $z=-1$.

By Cauchy’s theorem, the contour integral is zero. Thus, we have

$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = i 2 \pi \, e^{-i \pi p} \int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} + i e^{-i \pi p} \int_{-\pi}^{\pi} d\phi \left ( \log{\epsilon} + i \phi \right )$$

Note that

$$\begin{align}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} &= \int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} + O(\epsilon) \end{align} $$

and

$$\begin{align}\int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} &= \int_0^{1-\epsilon} dx \, x^p \left (\frac1x + \frac1{1-x} \right ) \\ &= \frac1p (1-\epsilon)^p – \left [\log{(1-x)} x^p \right ]_0^{1-\epsilon} + p \int_0^{1-\epsilon} dx \, x^{p-1} \log{(1-x)}\\ &= \frac1p – \log{\epsilon} + p \int_0^1 dx \, x^{p-1} \log{(1-x)} + O(\epsilon)\\ &= \frac1p – \log{\epsilon} – p \sum_{k=1}^{\infty} \frac1{k (k+p)}+ O(\epsilon) \\ &= \frac1p – \log{\epsilon}-\gamma -\psi(1+p)+ O(\epsilon) \\ &= – \log{\epsilon} – (\gamma + \psi(p))+ O(\epsilon) \end{align} $$

where $\psi$ is the digamma function. Note that the singular $\log{\epsilon}$ pieces cancel. The second piece of the second integral on the RHS vanishes as it is an odd function over a symmetric interval. Thus, we may take the limit as $\epsilon \to 0$ and we get

$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = -\frac{i 2 \pi \, e^{-i \pi p}}{1-e^{-i 2 \pi p}} (\gamma + \psi(p)) = -\frac{\pi}{\sin{\pi p}} (\gamma + \psi(p))$$

Alternatively, we may express this so that it is clear that the integral takes a positive value:

$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = \frac{\pi}{\sin{\pi p}} \left (\frac1p – H_p \right ) $$

where $H_p$ is the analytically continued harmonic number at $p$.

- Rain droplets falling on a table
- Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix?
- How to compute the topological space of fibered product of schemes?
- degree 3 Galois extension of $\mathbb{Q}$ not radical
- Deriving the mean of the Geometric Distribution
- Number of edge disjoint Hamiltonian cycles in a complete graph with even number of vertices.
- Determine and classify all singular points
- The sum of $n$ independent normal random variables.
- On the commutator subgroup of a group
- $\gcd(a,\operatorname{lcm}(b,c))=\operatorname{lcm}(\gcd(a,b),\gcd(a,c))$
- Convergence of alternating series based on prime numbers
- Continuous function on closed unit ball
- Let $f: \Bbb R \to [0,\infty)$ be a continuous function such that $g(x)=(f(x))^2$ is uniformly continuous. Which of the following is always true?
- Help to understand material implication
- Expected value of game involving 100-sided die