Handling Infinite Discontinuity in Multiple Integrals?

  • Context: Undergrad 
  • Thread starter Thread starter benorin
  • Start date Start date
  • Tags Tags
    Integral Multiple
Click For Summary
SUMMARY

The discussion centers on the handling of infinite discontinuity in multiple integrals, specifically regarding the Lerch Transcendent identity defined as $$\Phi (z,N,y) :=\sum_{q=0}^{\infty}\frac{z^q}{(q+y)^N}$$. The user seeks guidance on whether to set the upper bounds of integrals to $$1-\epsilon$$ or to use individual $$\epsilon_k$$ for each dimension as they approach zero. The identity holds under the conditions $$N\in\mathbb{Z}^+, |z|<1, y\in\mathbb{R}$$, but the user expresses uncertainty about the domains and the validity of their approach.

PREREQUISITES
  • Understanding of multiple integrals and their properties.
  • Familiarity with the Lerch Transcendent function and its applications.
  • Knowledge of limits in the context of calculus.
  • Basic concepts of complex analysis, particularly regarding the domain of complex variables.
NEXT STEPS
  • Research the properties and applications of the Lerch Transcendent function.
  • Study techniques for handling infinite discontinuities in integrals.
  • Learn about N-dimensional limits and their implications in calculus.
  • Review complex analysis focusing on the behavior of functions in the complex plane.
USEFUL FOR

Mathematicians, researchers in mathematical analysis, and students studying advanced calculus or complex analysis who are interested in multiple integrals and their convergence properties.

benorin
Science Advisor
Insights Author
Messages
1,442
Reaction score
191
TL;DR
I was writing a paper and derived some results using a theorem that turned out to be false, however some of these results are known to be true, so I am attempting to see if any of these are true, the one I mention here involves the Lerch Transcendent
The Lerch Transcendent identity from my paper which may or may not be true, for ##N\in\mathbb{Z}^+##, and I forget the domain of z and y, here it goes

$$\Phi (z,N,y) :=\sum_{q=0}^{\infty}\frac{z^q}{(q+y)^N}$$
$$=\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}\prod_{k=1}^{N}\left( \lambda_k^{y-1}\right)\left( 1-z\prod_{q=1}^{N}\lambda_q\right)^{-1}\, d\lambda_1d\lambda_2\cdots d\lambda_N$$

Some of the results I got using that untrue theorem were known results that were actually true, but I didn't check all of them, I wish to try to verify this identity by other means but I'll be honest I've not done much math for about 20 years so I need a little guidance here please?

How to handle the infinite discontinuity of the integrand at ##z=\lambda_k=1## for ##k=1,2,\ldots, N##? Do I take the upper bound of each integral to be ##1-\epsilon## and let ##\epsilon\rightarrow 0+##? Or do I have to set each upper bound to be ##1-\epsilon_k## and take a N-dimensional limit? Unsure how to start.
 
Physics news on Phys.org
For ##N\in\mathbb{Z}^+, |z|<1, y\in\mathbb{R}##, define

$$J=\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}\prod_{k=1}^{N}\left( \lambda_k^{y-1}\right)\left( 1-z\prod_{q=1}^{N}\lambda_q\right)^{-1}\, d\lambda_1d\lambda_2\cdots d\lambda_N$$
$$=\lim_{(\epsilon_1,\epsilon_2,\ldots,\epsilon_N )\rightarrow (0,0,\ldots, 0)}\int_{0}^{1-\epsilon_N}\int_{0}^{1-\epsilon_{N-1}}\cdots \int_{0}^{1-\epsilon_1}\sum_{q=0}^{\infty}z^q\prod_{k=1}^{N}\left( \lambda_k^{y+q-1}\right) \, d\lambda_1d\lambda_2\cdots d\lambda_N$$
$$=\lim_{(\epsilon_1,\epsilon_2,\ldots,\epsilon_N )\rightarrow (0,0,\ldots, 0)}\sum_{q=0}^{\infty}\frac{z^q}{(y+q)^N}\prod_{k=1}^{N}\left( 1-\epsilon_k\right) ^{y+q} =\sum_{q=0}^{\infty}\frac{z^q}{(q+y)^N} =:\Phi (z,N,y)$$

Conceivably one may take ##z,y\in\mathbb{C}## such that ##|z|<1##? My complex analysis is rusty to say the least, I can look it up later I guess. So my identity holds, just got to hammer out the domains. I used the N-dimensional limit as being the more general possibility, to cover all my bases. Was there any flaw in my work?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K