how to do integration for gamma function with limits

2 次查看（过去 30 天）

Maths Maths 2020-7-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/564656-how-to-do-integration-for-gamma-function-with-limits

评论： Walter Roberson 2020-7-25

clc
clear all
syms x
int(gamma(x),x,0,1)

9 个评论
显示 7更早的评论隐藏 7更早的评论

David Goodmanson 2020-7-15

在 MATLAB Online 中打开

gamma(z) = gamma(1+z)/z
Integral(gamma(z)) = Integral(gamma(z+1)/z -1/z + 1/z)
                   = Integral(gamma(z+1)-1)/z) + log(z)

Taking a look at the integrand numerator: Since gamma(1) = 1, (gamma(z+1)-1) is 0 when z = 0. Consequently (gamma(z+1)-1)/z equals a constant at z = 0. The integrand is bounded, so the integral can be integrated all the way down to small z with no problem. All the singular behavior is in the log term at the end. This says that

I = Integral{small,1} gamma(z) dz = C + log(z) |{small,1}
I = C -log(small)               % the result

where

C = Integral{small,1} (gamma(z+1)-1)/z)

is a slowly varying function of 'small'. So as a check it should be true that

I + log(small) = C % slowly varying

I can't integrate down as close to zero as Walter, but

C = integral(@(x) gamma(x),1e-23,1) +log(1e-23)
   C = -0.246695210286852
C = integral(@(x) gamma(x),1e-22,1) +log(1e-22)
   C = -0.246690599165717

Unfortunately, although the basic idea is demonstrated, the result for 1e-52 is

I = -log(1e-52) - .24669
   I =  119.4877

compared to Walter's 'about 121'.

Maths Maths 2020-7-25

I need to calculate the value of M can anyone help me to do this.

Walter Roberson 2020-7-25

It is not realistic that you would be able to find a closed form integral for that. You have t to unknown powers that are not guaranteed to be positive integers or even real-valued, and you have t^(-alp) where alp is an expression involving sin(t).

It is not even possible to find a closed form expression for int(t^sin(t), t, 0, 1)

请先登录，再进行评论。

请先登录，再回答此问题。