Complex Analysis: Integrate e^{\sin{z}} \cos{z} over Curve w_1 to w_2

[latentcorpse]

Let w_1,w_2 \in \mathbb{C} and \gamma be some smooth curve from w_1 to w_2.

Find \int_{\gamma} e^{\sin{z}} \cos{z} dz

this is holomorphic on the entire copmlex plan so we can't use a residue theorem. furthermore, we can't assume \gamma is a closed contour as we aren't told w_1=w_2 so it looks as if we're going to need to parameterise \gamma.

but we don't know what \gamma looks like. however we do know that any two point in the copmlex plane can be joined by a finite number of horizontal and vertical lines so if we use instead of \gamma a contour \gamma_1 \cup \gamma_2

where \gamma_1 is horizontal and \gamma_2 is vertical. this is my thoughts so far but parameterising these was still going to be pretty difficult so i decided to check if I am on the right lines or not. any advice?

 

[gabbagabbahey]

Hint: is this integral path independent? How can you tell?

 

[latentcorpse]

well the fundamental theorem of calculus gives that
for
\gamma:[a,b] \rightarrow U with \gamma(a)=\alpha,\gamma(b)=\beta
and f: U \rightarrow \mathbb{C} holomorphic on the open set U then, if \exists F'=f,

\int_{\gamma} f(z) dz = \int_{\gamma} F'(z) dz = \int_a^b F'(\gamma(t)) \gamma'(t) dt = \int_a^b (F \cdot \gamma(t))' dt = F \cdot \gamma(b) - F \cdot \gamma(a) = F(\beta)- F(\alpha)

i.e. if we can find such an F, the integral will depend only on the end point values and will therefore be path independent

for our case F=e^{\sin{z}} is such an F and so the answer is

\int_{\gamma} e^{\sin{z}} \cos{z} dz = e^{\sin{w_2}} - e^{\sin{w_1}}=e^{\sin{w_2}-\sin{w_1}}

is that ok?

is there an easier way to tell its path independent?

also, can you recommend a good book that has plenty of worked exapmles on integrals like this (and stuff using Cauchy residue theorem etc)?

 

[latentcorpse]

also, how did you realize to argue about path independence?

 

[yyat]

A holomorphic function defined on \mathbb{C} always has an antiderivative (for example compute the Taylor series and integrate each term).

As you pointed out, existence of an antiderivate implies independence of path for the integral.

More generally, suppose f:D\to\mathbb{C} is a holomorphic function and \alpha, \beta are paths with the same endpoints. If \alpha can be continuously deformed in D into \beta while keeping the endpoints fixed, then the integral of f over the two paths is the same. This follows easily from Cauchy's integral theorem.

 

[gabbagabbahey]

\int_{\gamma} e^{\sin{z}} \cos{z} dz = e^{\sin{w_2}} - e^{\sin{w_1}}=e^{\sin{w_2}-\sin{w_1}}

is that ok?

Aside from your final answer being in grievous error...sure

is there an easier way to tell its path independent?

Like yyat said, whenever the integrand is holomorphic, the integral is path independent.

also, how did you realize to argue about path independence?

The question wouldn't have made much sense if the integral depended on the path now would it?

 

[latentcorpse]

yeah. the final answer step was a BAD mistake.
surely then we could answer by this means for any given function but here it's particularly useful because we were given no information about the path itself other than the endpoints