Solve a very complex line integral

[Mathman23]

Homework Statement

Solve $$I = \int_{\gamma} f(z) dz$$ where $$\gamma(t) = e^{i \cdot t}$$ and $$0 \leq t \leq \pi$$

Homework Equations

Do I use integration by substitution??

The Attempt at a Solution

If I treat this as a line-integral I get:

$$I = \int_{a}^{b} f(\gamma(t)) \cdot \gamma'(t) dt = \int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt$$

then If I choose

$$u = e^{it}$$ and $$dt = i \cdot e^{i \cdot u} du$$

where $$t=0, u=e^{i \cdot 0} = 1$$ and $$t=\pi, u=e^{i \cdot 0} = -1$$

$$\int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt = \int_{1}^{-1} (i \cdot (cos(u) \cdot u) du = (i \cdot (cos(u) + u \cdot sin(u))) \cdot i \cdot e^{i \cdot u} ]_{1}^{-1} = 2 sin(1) \cdot ((cos(1) + sin(1)) \cdot i$$

Am I on the right path here??

Best regards
Fred

 

[HallsofIvy]

Homework Statement

Solve $$I = \int_{\gamma} f(z) dz$$ where $$\gamma(t) = e^{i \cdot t}$$ and $$0 \leq t \leq \pi$$

Homework Equations

Do I use integration by substitution??

The Attempt at a Solution

If I treat this as a line-integral I get:

$$I = \int_{a}^{b} f(\gamma(t)) \cdot \gamma'(t) dt = \int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt$$

So f(z)= cos(z)? You didn't mention that before! Cosine is an analytic function isn't it?

then If I choose

$$u = e^{it}$$ and $$dt = i \cdot e^{i \cdot u} du$$

where $$t=0, u=e^{i \cdot 0} = 1$$ and $$t=\pi, u=e^{i \cdot 0} = -1$$

$$\int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt = \int_{1}^{-1} (i \cdot (cos(u) \cdot u) du = (i \cdot (cos(u) + u \cdot sin(u))) \cdot i \cdot e^{i \cdot u} ]_{1}^{-1} = 2 sin(1) \cdot ((cos(1) + sin(1)) \cdot i$$

Am I on the right path here??

Best regards
Fred

That might work but looks overly complicated! How about just finding the anti-derivative of cos(z) (I'm sure you know it!) and evaluating it at the two endpoints of the contour?

 

[Mathman23]

Hi there,

You are right :)

Then I get

$$<br /> \int_0^{\pi}Cos[e^{it}]ie^{it}dt$$

and I let $$u=e^{it}$$

then:

$$du=ie^{it}dt$$

right?

so that part is already in the integral so now it's just:

$$<br /> \int_0^{\pi}Cos[e^{it}]ie^{it}dt=\int_1^{-1}Cos<u>du</u>$$

right? I then get $$-2Sin[1]$$

I got that result in Maple too. Had an idear that it was the right one :)

By following the method above. I get:

1)$$f(z) = sinh(z)$$ and 2) $$f(z) = (exp(z))^3$$ and 3)$$f(z) = tan(z)$$.

1) Solution

$$\int_{0}^{\pi} sinh(e^{it}) \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} sinh(u) du = 0$$

2) Solution

$$\int_{0}^{\pi} (e^{e^{it}})^3 \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} (e^{u})^3 du = \frac{-2 sinh(3)}{3}$$

3) Solution

$$\int_{0}^{\pi} tan(e^{it}) \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} tan(u) du = 0$$

How do they look? Am I following you guidelines correctly??

Best regards Fred

 

[Office_Shredder]

$$dt = i \cdot e^{i \cdot u} du$$

This should be $$du = i \cdot e^{i \cdot t} dt$$

I think...

notice your integral becomes easier