matematikk.net

Hei sitter her og prøver å løse et integrasjonsstykke. Lang Jule ferie osv sliter, kan noen hjelpe meg. På forhånd takk

[itgl][/itgl]1/x*e^-x

er du sikker på at du vil løse dette integralet?
Integralet gir Ei(-x) som er litt vanskelig å forklare.
Eller du må skrive integranden som rekke.

nettopp, var også på denne adressen. Der får du Ei(-x) som svar.

Andrina skrev:nettopp, var også på denne adressen. Der får du Ei(-x) som svar.

Hehhe, bruker mathematica jeg

Uansett - hva vil Ei(-x) si?

Hva sier mathematica da?

dette er definisjonen av Ei som den står hos Maple:

Ei - The Exponential Integral
Calling Sequence
Ei(z)
Ei(a, z)
Parameters
z - algebraic expression
a - algebraic expression
Description
The exponential integrals, Ei(a, z), are defined for Re(z) > 0 by
> Ei(a, z) = convert(Ei(a, z), Int) assuming Re(z) > 0;
/infinity
| (-a)
Ei(a, z) = | exp(-_k1 z) _k1 d_k1
|
/1
This classical definition is extended by analytic continuation to the entire complex plane using
> Ei(a, z) = z^(a-1)*GAMMA(1-a, z);
(a - 1)
Ei(a, z) = z GAMMA(1 - a, z)
with the exception of the point 0 in the case of Ei(1, z).
For all of these functions, 0 is a branch point and the negative real axis is the branch cut. The values on the branch cut are assigned such that the functions are continuous in the direction of increasing argument (equivalently, from above).
The classical definition for the 1-argument exponential integral is a Cauchy Principal Value integral, defined for real arguments x, as the following
> convert(Ei(x),Int) assuming x::real;
/x
| exp(_k1)
| -------- d_k1
| _k1
/-infinity
> value(%);
Ei(x)
for x < 0, Ei(x) = -Ei(1, -x). This classical definition is extended to the entire complex plane using
Ei(z) = -Ei(1, -z) + (ln(z) - ln(1/z))/2 - ln(-z)
Note that this extension has its branch cut on the negative real axis, but unlike for the 2-argument Ei functions this extension is not continuous onto the branch cut from either above or below. That is, this extension provides an analytic continuation of Ei(z) from the positive real axis, but not in any direction from the negative real axis. If you want a continuation from the negative real axis, use -Ei(1, -z) in place of Ei(z).
Examples
> Ei(1,1.);
0.2193839344
> Ei(1,-1.);
-1.895117816 - 3.141592654 I
> expand(Ei(3,x));
1 1 1 2
- exp(-x) - - x exp(-x) + - x Ei(1, x)
2 2 2
> simplify(Ei(1,I*x)+Ei(1,-I*x));
-2 Ci(x) - I Pi + I Pi csgn(x)
> Ei(5, 3+I);
Ei(5, 3 + I)
> evalf(%);
0.002746760454 - 0.006023680639 I
> Ei(1.);
1.895117816
> Ei(1.+0.*I);
1.895117816 + 0. I
> Ei(1.-0.*I);
1.895117816 + 0. I
> Ei(-1.);
-0.2193839344
> Ei(-1.+0.*I);
-0.2193839344 + 3.141592654 I
> Ei(-1.-0.*I);
-0.2193839344 - 3.141592654 I
> Ei(1.3+4.7*I);
-0.7490731390 + 3.097526006 I
> int(exp(-3*t)/t, t=-x..infinity, CauchyPrincipalValue);
-Ei(3 x)