Integrals for the LogGamma & polygamma fcns

  • Thread starter benorin
  • Start date
  • Tags
    Integrals
In summary, the conversation discusses proving a formula for the polygamma functions by differentiating under the integral sign multiple times. It is shown that the integral is absolutely convergent, which is necessary for applying the differentiation rule, by bounding the integrand and using the comparison test.
  • #1
benorin
Homework Helper
Insights Author
1,435
186
I have proved that for [tex]\Re [z]>0,[/tex]

[tex]\log \Gamma (z) = \int_{0}^{\infty} \left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}dt[/tex]​

and I wish to justify differentiating under the integral sign (n+1)-times to give integral formulas for the polygamma functions. Let

[tex]f(t)=\left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}[/tex]​

I have proved that if [tex]\Re [z]>0,\mbox{ and }|z|<\infty,[/tex] then

[tex]\lim_{t\rightarrow 0^+} f(t)=\frac{1}{2}z^2-\frac{3}{2}z+1,\mbox{ and }\lim_{t\rightarrow\infty} f(t)=0[/tex]​

and if I recall correctly, it is sufficient that the integral in question be absolutely convergent to apply the differentiation rule.

EDIT: So how do I absolute convergence? Any hints?
 
Last edited:
Physics news on Phys.org
  • #2
To show absolute convergence, you need to show that the integral converges absolutely. This means that it is bounded in absolute value, meaning that ∣∫f(t)dt∣<∞. You can do this by showing that the integrand is bounded in absolute value, meaning that ∣f(t)∣<M for some constant M. To do this, start with the term inside the parentheses in f(t). Since e−t is always positive and less than 1, the denominator of the fraction is always greater than 0, so the numerator is bounded in absolute value by |1−e−(z−1)t|≤2. The rest of the terms in the integrand are all either constants or monotonically decreasing functions of t, so ∣f(t)∣≤C⋅2/t, where C is a constant. Now use comparison test to show that the integral converges absolutely.
 

1. What is the LogGamma function?

The LogGamma function is a special mathematical function that is used to calculate the natural logarithm of the Gamma function. It is denoted as ln(Γ(x)) and is defined as the integral of the logarithm of the Gamma function over the real numbers.

2. How is the LogGamma function used in science?

The LogGamma function is commonly used in various scientific fields, such as mathematics, statistics, and physics. It is often used to simplify complicated calculations involving the Gamma function and to solve certain types of differential equations.

3. What is the relationship between the LogGamma and polygamma functions?

The polygamma function is the derivative of the LogGamma function. In other words, the polygamma function is the n-th derivative of the LogGamma function. The LogGamma function and the polygamma function are closely related and are often used together in mathematical calculations.

4. How do we evaluate integrals involving the LogGamma and polygamma functions?

Evaluating integrals involving the LogGamma and polygamma functions can be challenging and often requires advanced mathematical techniques. One approach is to use numerical methods, such as the trapezoidal rule or Simpson's rule, to approximate the integral. Another approach is to use specialized software or computer programs that can calculate the integral accurately.

5. Can the LogGamma and polygamma functions be extended to complex numbers?

Yes, both the LogGamma and polygamma functions can be extended to complex numbers. The resulting functions are known as the complex LogGamma and polygamma functions. These complex functions have important applications in fields such as quantum mechanics and number theory.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
313
  • Calculus and Beyond Homework Help
Replies
6
Views
397
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
489
  • Calculus and Beyond Homework Help
Replies
3
Views
429
  • Calculus and Beyond Homework Help
Replies
1
Views
531
  • Calculus and Beyond Homework Help
Replies
17
Views
450
  • Calculus and Beyond Homework Help
Replies
7
Views
623
  • Calculus and Beyond Homework Help
Replies
6
Views
158
  • Calculus and Beyond Homework Help
Replies
2
Views
836
Back
Top