Explain this method for integrals (complex analysis)

In summary: When he integrated the digamma function, he got the residues at -2, -1, 0, and the sum of all remaining residues (all at positive integers). The sum of all components is zero. As the sum contains ∑2Hn(n+1)(n+2)3\sum \frac {2 H_n}{(n+1)(n+2)^3} this allows to solve the equation for this expression. You can then use that equation for the original problem.
  • #1
412
1
I saw this method of calculating:

$$I = \int_{0}^{1} \log^2(1-x)\log^2(x) dx$$

http://math.stackexchange.com/questions/959701/evaluate-int1-0-log21-x-log2x-dx

Can you take a look at M.N.C.E.'s method?

I don't understand a few things.

Somehow he makes the relation:

$$\frac{4H_n}{(n+1)(n+2)^3} = \frac{\left( \gamma + \psi(-z) \right)^2}{(z+1)(z+2)^3}$$

How is this established?

And this I don't understand, why did he integrate it,?

And then after he states: "At the positive integers," what is he doing with the residues. I know the residue theorem etc, but I don't understand what he is exactly doing?
Thanks
 
Physics news on Phys.org
  • #2
That relation is not made. The logic:

Consider this function f(x) someone made up: if we integrate it over a square (as described in the comments) the integral is zero. Another way to express the same integral is via its residues: their sum has to be zero.
He then calculates the residues at -2, -1, 0 and the sum of all remaining residues (all at positive integers). The sum of all components is zero. As the sum contains ##\sum \frac {2 H_n}{(n+1)(n+2)^3}## this allows to solve the equation for this expression. You can then use that equation for the original problem.
 
  • #3
mfb said:
That relation is not made. The logic:

Consider this function f(x) someone made up: if we integrate it over a square (as described in the comments) the integral is zero. Another way to express the same integral is via its residues: their sum has to be zero.
He then calculates the residues at -2, -1, 0 and the sum of all remaining residues (all at positive integers). The sum of all components is zero. As the sum contains ∑2Hn(n+1)(n+2)3\sum \frac {2 H_n}{(n+1)(n+2)^3} this allows to solve the equation for this expression. You can then use that equation for the original problem.

@mfb, thankyou very much, this was excellent help, physicsforums is great. I would like to ask a few things if you don't mind.

Question 1) He does not define the square though? Where are the vertices?
-I will assume that (-2, -1, 0) are INSIDE the rectangle, not on or outside.

Question 2) How are there infinite singularities at positive integers? http://m.wolframalpha.com/input/?i=digamma(-z)&x=0&y=0
I think I know. \digamma(-z) yields complex infinity for all positive integer values of z.
So there are non removable singularities.

Question 3) How does he get this:

$$\sum_{n=1}^{\infty} Res(f, n) = \sum_{n=1}^{\infty} Res_{z=n} \frac{1}{(z+1)(z+2)^3(z-n)^2} + \frac{2H_n}{(z+1)(z+2)^3(z-n)}$$

More specifically, why does he have (z-n) in the denominator? And where did the $H_n$ appear from?

Thats all for now, hopefully you can guide me, thanks.
 
  • #4
Amad27 said:
He does not define the square though?
It is described in the smaller comments below the main post.

I don't know how he got those residuals. Must come from the digamma function.
 
  • #5
mfb said:
It is described in the smaller comments below the main post.

I don't know how he got those residuals. Must come from the digamma function.
Okay. mm.. I don't understand.

When he integrated the digamma function, how does the result in a function with H_n?? Thanks
 

1. What is complex analysis and why is it used for integrals?

Complex analysis is a branch of mathematics that deals with complex numbers, which are numbers that contain both a real and an imaginary component. It is used for integrals because it allows for the integration of functions that cannot be solved using traditional methods, such as those involving trigonometric or logarithmic functions.

2. How does complex analysis differ from real analysis?

Complex analysis differs from real analysis in that it involves the study of complex-valued functions, while real analysis deals with real-valued functions. In complex analysis, the concept of differentiation and integration is extended to complex numbers, which have both real and imaginary parts, while real analysis only deals with real numbers.

3. What is the Cauchy-Riemann equations and how does it relate to complex analysis and integrals?

The Cauchy-Riemann equations are a set of conditions that must be satisfied for a function to be differentiable in the complex plane. They relate to complex analysis and integrals because they provide a way to determine if a function is analytic, meaning it can be expressed as a power series, and thus can be integrated using complex analysis techniques.

4. Can complex analysis be used to evaluate real integrals?

Yes, complex analysis can be used to evaluate real integrals. This is because real integrals can often be transformed into complex integrals, which can then be solved using complex analysis techniques. This method is known as the method of contour integration.

5. Are there any practical applications of complex analysis for integrals?

Yes, complex analysis has many practical applications for integrals. It is used in a variety of fields, such as physics, engineering, and economics, to solve complex problems that cannot be solved using traditional methods. It is also used in the development of algorithms for numerical integration, which are used in computer simulations and data analysis.

Similar threads

Replies
7
Views
1K
Replies
4
Views
645
Replies
4
Views
1K
  • Topology and Analysis
Replies
2
Views
499
  • Topology and Analysis
Replies
4
Views
143
  • Topology and Analysis
Replies
2
Views
1K
Replies
6
Views
725
  • Topology and Analysis
Replies
1
Views
278
  • Topology and Analysis
Replies
5
Views
725
  • Topology and Analysis
Replies
2
Views
1K
Back
Top