Never ending integration by parts

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
Smed
Messages
36
Reaction score
1

Homework Statement


[tex]\int_0^\infty{ \frac{1}{x} e^{-x}}[/tex]

Homework Equations


Integration by parts
[tex]\int{u dv} = uv - \int{v du}[/tex]

The Attempt at a Solution


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

[tex]-\frac{1}{x} e^{-x} - \int_0^\infty{-e^{-x} \frac{1}{x^2}} dx[/tex]
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.
 
Physics news on Phys.org
Smed said:

Homework Statement


[tex]\int_0^\infty{ \frac{1}{x} e^{-x}}[/tex]

Homework Equations


Integration by parts
[tex]\int{u dv} = uv - \int{v du}[/tex]

The Attempt at a Solution


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

[tex]-\frac{1}{x} e^{-x} - \int_0^\infty{-e^{-x} \frac{1}{x^2}} dx[/tex]
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).