òex/x
Editor: This integral often crops up in Ask NRICH. Similar
integrals also frequently arise which can be converted via integration by
parts and substitution into this form are òex logx and òlogx /ex.]
By Arun Iyer (P4587) on Friday, June 08,
2001 - 07:48 pm:
What is the value of
the indefinite integral òex/x?
By Brad Rodgers (P1930) on Friday, June
08, 2001 - 11:17 pm:
The answer is essentially, that there is no answer. The integral of this
function cannot be expressed by any combination of algebraic functions (though
I'm afraid I don't really know how to prove this). Instead, mathematicians
invented the function Ei(x), defining it as
| Ei(x)=g+ln(x)+ |
¥
å
n=1
|
xn /(n!n)
|
(with the g on there for convenience in other calculations;
| g = |
lim
n®¥
|
]( |
n
å
k=1
|
(1/k)-ln(n))
|
)
To show that òex /x=Ei(x), we want to show that d/dx(Ei(x))=ex /x.
So,
| d/dx(Ei(x))=d/dx(g)+d/dx(ln(x))+d/dx( |
¥
å
n=1
|
xn /(n!× n)
|
=1/x+1+x/2!+x2 /3!+x3 /4!
(i.e.
|
¥
å
n=1
|
xn /(n!×n)=x+x2 /(2!)+x3 /(3!×3)+ x4 /(4!×4)+¼
|
So,
| d/dx( |
¥
å
n=1
|
xn /(n!×n))=d/dx(x)+d/dx(x2 /2!×2)+ d/dx(x3 /3!×3)+¼)
|
As, recalling the power series for ex,
ex /x=1/x+1+x/2!+x2/3!+¼
we now have shown that d/dx(Ei(x))=ex/x, as required.
If there's anything you don't understand, don't hesitate to write
again,
Brad
By Arun Iyer (P4587) on Thursday, June 14,
2001 - 07:47 pm:
Thanks Brad,
I think I get the picture that you are trying to show me.
Arun