Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Jun 26, 2017 5:42:48 GMT
Vasco,
Have you done anything with this?
p. 410, Para. 3, use the Fundamental Theorem to evaluate the integral for $e^z$. By equating the imaginary part of your answer with the imaginary part of the parametric evaluation, deduce that
$\hspace{5em}\int_0^1(2 t cos t^2) e^t dt = e sin 1$.
Gary
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Post by Admin on Jun 26, 2017 5:54:57 GMT
Vasco, Have you done anything with this? p. 410, Para. 3, use the Fundamental Theorem to evaluate the integral for $e^z$. By equating the imaginary part of your answer with the imaginary part of the parametric evaluation, deduce that
$\hspace{5em}\int_0^1(2 t cos t^2) e^t dt = e sin 1$.
GaryGary No, but it looks different in my revision of the book: $\hspace{5em}\int_0^1(2 t cos t^2+sin t^2) e^t dt = e sin 1$ Vasco PS ...and it seems to work out OK in this form. |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 26, 2017 14:16:59 GMT
Vasco, Have you done anything with this? p. 410, Para. 3, use the Fundamental Theorem to evaluate the integral for $e^z$. By equating the imaginary part of your answer with the imaginary part of the parametric evaluation, deduce that
$\hspace{5em}\int_0^1(2 t cos t^2) e^t dt = e sin 1$.
GaryGary No, but it looks different in my revision of the book: $\hspace{5em}\int_0^1(2 t cos t^2+sin t^2) e^t dt = e sin 1$ Vasco PS ...and it seems to work out OK in this form. Vasco, My book is the same. I just mis-keyed when I created the message. I was able to do the previous four suggested exercises without difficulty (I think), but this one eludes me. I started with $z(t) = r(cos t + i sin t)$ leading to the integral $r[\int_0^1(e^{r(cos t + i sin t)}(- sin t) dt + \int_0^1(e^{r(cos t + i sin t)}(cos t) dt]$ but my skills at integration don't cover this. I tried simplifying to $z(t) = 1 + e^t$ but that left me without an imaginary part. Gary |
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Post by Admin on Jun 26, 2017 15:44:25 GMT
Gary
Here is my version. The solution is a bit simpler than you might expect.
We have to integrate $z$ along the contour $z(t)=t+it^2, (0\leq t\leq 1)$
Using the Fundamental Theorem we have
$\int_0^{1+i}e^zdz=[e^{1+i}-e^0]=ee^i-1=e(\cos 1+i\sin 1)-1=e\cos 1-1+ie\sin 1$
Using parametric evaluation we have
$\int_{t=0}^{t=1}e^zdz=\int_0^1e^{t+it^2}vdt=\int_0^1e^{t+it^2}(1+2it)dt=\int_0^1e^t(e^{it^2}+2ite^{it^2})dt$
Equating the imaginary parts of each of these we find
$\int_0^1(2t\cos t^2+\sin t^2)e^tdt=e\sin 1$.
Of course by equating real parts we could get another result.
Vasco
PS Notice that we don't evaluate the parametric integral because if we did, we would just get $e\sin 1$,.
Also note that the integral of a complex function is the sum of the integral of the real and imaginary parts of the function separately.
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 26, 2017 23:34:42 GMT
Gary Here is my version. The solution is a bit simpler than you might expect. We have to integrate $z$ along the contour $z(t)=t+it^2, (0\leq t\leq 1)$ Using the Fundamental Theorem we have $\int_0^{1+i}e^zdz=[e^{1+i}-e^0]=ee^i-1=e(\cos 1+i\sin 1)-1=e\cos 1-1+ie\sin 1$ Using parametric evaluation we have $\int_{t=0}^{t=1}e^zdz=\int_0^1e^{t+it^2}vdt=\int_0^1e^{t+it^2}(1+2it)dt=\int_0^1e^t(e^{it^2}+2ite^{it^2})dt$ Equating the imaginary parts of each of these we find $\int_0^1(2t\cos t^2+\sin t^2)e^tdt=e\sin 1$. Of course by equating real parts we could get another result. Vasco PS Notice that we don't evaluate the parametric integral because if we did, we would just get $e\sin 1$,. Also note that the integral of a complex function is the sum of the integral of the real and imaginary parts of the function separately. Vasco, Thank you. My problem was I didn't realize that the problem was to integrate $e^z$ along the parabola. I'll try it again without further inspection and then reread your comment. I have done the evaluations and posted in my notes for Chapter 8. It's an interesting notion to evaluate the integral of $e^z$ over a parabola or other interesting curve (or surface!). Can you think of a real world application? Gary |
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Post by Admin on Jun 27, 2017 7:50:46 GMT
Gary
The real-world applications are many and some are explored in the later chapters, 10 - 12. If you look on the internet you'll find plenty of examples. If we find any really good sites we could put some links in the forum as we did with the pseudosphere etc.
Vasco
PS If you were thinking particularly of the function $e^z$, then none come immediately to mind.
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