Never ending integration by parts

[Smed]

Homework Statement

\int_0^\infty{ \frac{1}{x} e^{-x}}

Homework Equations

Integration by parts
\int{u dv} = uv - \int{v du}

The Attempt at a Solution

u = \frac{1}{x}
du = \frac{1}{x^2} dx
v = -e^{-x}
dv = e^{-x} dx

-\frac{1}{x} e^{-x} - \int_0^\infty{-e^{-x} \frac{1}{x^2}} dx
It looks like this process is going to go on forever because I can't get rid of the 1/x term. Could someone please give some guidance on how this is done? Thanks.

 

[Dick]

Homework Statement

\int_0^\infty{ \frac{1}{x} e^{-x}}

Homework Equations

Integration by parts
\int{u dv} = uv - \int{v du}

The Attempt at a Solution

u = \frac{1}{x}
du = \frac{1}{x^2} dx
v = -e^{-x}
dv = e^{-x} dx

-\frac{1}{x} e^{-x} - \int_0^\infty{-e^{-x} \frac{1}{x^2}} dx
It looks like this process is going to go on forever because I can't get rid of the 1/x term. Could someone please give some guidance on how this is done? Thanks.

You can't reduce it to a simple function using integration by parts. The integral defines a special function called the 'exponential integral', Ei(x).

 

[obafgkmrns]

Actually, the exponential integral Ei(x) is defined a little differently. See http://mathworld.wolfram.com/ExponentialIntegral.html for example. In any case, given your limits of zero to infinity, the integral diverges.

 

[Dick]

Actually, the exponential integral Ei(x) is defined a little differently. See http://mathworld.wolfram.com/ExponentialIntegral.html for example. In any case, given your limits of zero to infinity, the integral diverges.

Ooops. Thanks for the correction. I didn't pay any attention to the fact it was a definite integral.