Integral Help exp (-x)/x with 0 to t limits

[piyush]

I am trying to get this integral but no clues about how to proceed
\int exp(-x)/x dx

with limits from 0 to t ( any real number, non infinity)

I know of the limits are from t to infinity, there is a standard integral known as Ei(x). please help/share ideas in solving this!


Regards and Thanks

Piyush

[meister]

Mathematica doesn't like it, it tosses Ei[-x] back at me, whatever that is. Not sure if there's an analytical solution.

[Zurtex]

I'm not sure it is valid at x=0. If we looks at the infinite series:

\frac{e^{-x}}{x} = \frac{1}{x} - \sum_{n=0}^{\infty} (-1)^n\frac{x^n}{(n+1)!}

Therefore:

\int \frac{e^{-x}}{x} dx = \ln x - \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n+1}}{(n+1)!*(n+1)} + C

This is clearly not valid for x = 0, which kind of makes sense as e^{-x}/x is not valid at x = 0.

[Gokul43201]

Since 1/x is non-integrable in any interval that includes 0 (and exp(0)=1) since the integral diverges, so would be (1/x)*exp(-x).

ie. the integral diverges to infinity.

[piyush]

Thanks guys: I have a few questions on this:

1. If I integrate within limits say 0.001 to t, is it going to introduce a lot of error in subsequent calculations that i carry out?

2. Actually I am trying to obtain this integral


\int f(r,x)dx with limits 0 to t
where f(r,x) is the Laplacian inverse of F(r, s) where F(r,s) is

\frac{K_{o}(r\sqrt{s})}{\sqrt{s}K_{1}(\sqrt{s})}

Ko and K1 are modified Bessel functions of zero and first order
Some simplifications resulted in the integral that I had put initially..

Can anyone please suggest a way out?

[kuahji]

If we know t>0 then what is the best way to integrate this problem, without computer help?