The Upper Incomplete Gamma Function

[nasi112]

The upper incomplete gamma function is defined as

[MATH]\Gamma(s,x) = \int_{x}^{\infty} t^{s-1} \ e^{-t} \ dt[/MATH]
Now, I want to solve this integral.

[MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx = \Gamma (0,\ln(2)) = \int_{\ln 2}^{\infty} t^{-1} \ e^{-t} \ dt = -\gamma - \ln(\ln(2)) - \sum_{k=1}^{\infty} \frac{(-\ln(2))^k}{k(k!)} \approx 0.378671043061088[/MATH]
where [MATH]\gamma[/MATH] is Euler-Mascheroni constant [MATH]\approx 0.5772156649015328606065120900824024310421 [/MATH]
Why the integral gave gamma function [MATH]\Gamma(s,x)[/MATH], the parameters, [MATH]s = 0 [/MATH] and [MATH]x = \ln(2)[/MATH]?

 

[Deleted member 4993]

The upper incomplete gamma function is defined as

[MATH]\Gamma(s,x) = \int_{x}^{\infty} t^{s-1} \ e^{-t} \ dt[/MATH]
Now, I want to solve this integral.

[MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx = \Gamma (0,\ln(2)) = \int_{\ln 2}^{\infty} t^{-1} \ e^{-t} \ dt = -\gamma - \ln(\ln(2)) - \sum_{k=1}^{\infty} \frac{(-\ln(2))^k}{k(k!)} \approx 0.378671043061088[/MATH]
where [MATH]\gamma[/MATH] is Euler-Mascheroni constant [MATH]\approx 0.5772156649015328606065120900824024310421 [/MATH]
Why the integral gave gamma function [MATH]\Gamma(s,x)[/MATH], the parameters, [MATH]s = 0 [/MATH] and [MATH]x = \ln(2)[/MATH]?

Simple substitution:

2^(-x) = e^(-t)

 

[nasi112]

Simple substitution:

2^(-x) = e^(-t)

I know this substitution.

I am asking why [MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx = \Gamma (0,\ln(2))[/MATH].

I will give you a few examples, and you will see what I mean.

[MATH]\int_{1}^{\infty} \frac{1}{x3^x} \ dx = \Gamma (0,\ln(3))[/MATH]
[MATH]\int_{1}^{56} \frac{1}{x3^x} \ dx = \Gamma (0,56\ln(3))[/MATH]
[MATH]\int_{9}^{15} \frac{1}{x8^x} \ dx = \Gamma (0,9\ln(8)) - \Gamma (0,15\ln(8))[/MATH]
When you get this form of integral, gamma function allows you to plug in numbers directly inside it. Why?

What is the process being done here?

Is there an explanation for that step with some kind of infinite series such as Taylor, Fourier, or other?

 

[lex]

[MATH]\int_{1}^{56} \frac{1}{x3^x} \ dx = \Gamma (0,56\ln(3))[/MATH]

?

 

[Deleted member 4993]

I know this substitution.

I am asking why [MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx = \Gamma (0,\ln(2))[/MATH].

I will give you a few examples, and you will see what I mean.

[MATH]\int_{1}^{\infty} \frac{1}{x3^x} \ dx = \Gamma (0,\ln(3))[/MATH]
[MATH]\int_{1}^{56} \frac{1}{x3^x} \ dx = \Gamma (0,56\ln(3))[/MATH]
[MATH]\int_{9}^{15} \frac{1}{x8^x} \ dx = \Gamma (0,9\ln(8)) - \Gamma (0,15\ln(8))[/MATH]
When you get this form of integral, gamma function allows you to plug in numbers directly inside it. Why?

What is the process being done here?

Is there an explanation for that step with some kind of infinite series such as Taylor, Fourier, or other?

Please show us what does:

[MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx [/MATH].

transform to,

when you substitute:

2^-x = e^-t

 

[nasi112]

lol, you are Very Accurate

it should be

[MATH]\int_{1}^{56} \frac{1}{x3^x} \ dx = \Gamma (0,\ln(3)) - \Gamma (0,56\ln(3))[/MATH]

 

[nasi112]

Please show us what does:

[MATH]\int_{1}^{\infty} \frac{1}{x2^x} \ dx [/MATH].

transform to,

when you substitute:

2^-x = e^-t

it transforms to

[MATH]\int_{\ln 2}^{\infty} \frac{1}{t e^t} \ dt[/MATH]
OK, this has answered why there is [MATH]\ln(2)[/MATH] in [MATH]\Gamma(0,\ln(2))[/MATH]
But what about the zero?

 

[Deleted member 4993]

it transforms to

[MATH]\int_{\ln 2}^{\infty} \frac{1}{t e^t} \ dt[/MATH]
OK, this has answered why there is [MATH]\ln(2)[/MATH] in [MATH]\Gamma(0,\ln(2))[/MATH]
But what about the zero?

[MATH]\int_{\ln 2}^{\infty} \frac{1}{t e^t} \ dt[/MATH]
=[MATH]\int_{\ln 2}^{\infty} t^{0-1} e^{-t}\ dt[/MATH]
Now think ..........

 

[nasi112]

Oh, it is so obvious that [MATH]s = 0[/MATH]. I don't know why I didn't see it. Lol

Maybe lack of sleep? I think that I am losing my skills?

Thanks Prof Khan.