Solving closed-form solution of integral by Matlab!!!

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kader
kader 2016 年 12 月 15 日
コメント済み: Walter Roberson 2016 年 12 月 15 日
Let a, b, and c are constants. How to solve the closed-form solution of the following integral by Matlab from 0 to infinity
1/(1+y)*1/(a+y)*1/(b+y)*exp(-cy)
Thanks in advance for your kind cooperation
  3 件のコメント
Walter Roberson
Walter Roberson 2016 年 12 月 15 日
Are there any constraints on the constants? Real-valued? Non-negative?
Walter Roberson
Walter Roberson 2016 年 12 月 15 日
Under the assumption that none of the values are negative,
(exp(c*a)*(b-1)*Ei(1, c*a)-exp(c*b)*(a-1)*Ei(1, c*b)+exp(c)*Ei(1, c)*(a-b))/((b-1)*(a-1)*(a-b))
where Ei is the Exponential Integral

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回答 (1 件)

David Goodmanson
David Goodmanson 2016 年 12 月 15 日
Hi kader, This depends on what you consider to be closed form. I assume the factor in front is 1/[(1+y)(a+y)(b+y)]. You can solve for constants F,A,B in partial fractions
1/((1+y)(a+y)(b+y)) = F/(1+y) +A/(a+y) +B/(b+y)
Then e.g,
A Int{0,inf} exp(-cy)/(a+y) dy
= A exp(ac) Int{ac,inf} exp(-z)/z dz
= A exp(ac) E1(ac)
where E1 is the exponential integral, which Matlab has.

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