Integration

👨🏻‍🏫 學術台數學

[匿名]

2007-04-04 7:55 pm

Do any one know how to do the following integrtion?

Indefinite Integration {[e(x)]/x}dx

更新1:

that's means Integration {[e^(x)] / x}dx = {[e^(x)] / x} Summation i = 0 to infinity {i! / [x^(i)]} so, what is the final answer? I don't know the limit of this infinite series

更新2:

So, does the limit exists or not, in your final answer? However, I have found another approximation method by Taylor Series first-order approximation

回答 (2)

匿名

2007-04-04 8:20 pm

✔ 最佳答案

Indefinite Integration {[e^(x)]/x}dx

int [P(x)e^(ax)]dx = P(x)e^(ax)/a - P'(x)e^(ax)/(a^2) + P"(x)e^(ax)/(a^3) -...
Let P(x)=1/x, a=1
int {[e^(x)]/x}dx = e^(x)/x - e^(x)*(-x^(-2))
=e^(x)/x + e^(x)/(x^2)
or = [(x+1)e^(x)]/(x^2)

參考: myself

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HaHa

2007-04-04 11:43 pm

其實答案是一個叫做exponential integral的東西
你可以參考這個網站
http://mathworld.wolfram.com/ExponentialIntegral.html

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