Loser66
  • Loser66
integral help \[\int_a^1 \dfrac{e^{(i-1)/t}{t^2}dt\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
|dw:1446913995921:dw|
Loser66
  • Loser66
@thomas5267
thomas5267
  • thomas5267
i is imaginary number?

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Loser66
  • Loser66
yes
thomas5267
  • thomas5267
u=(i-1)/t substitution?
thomas5267
  • thomas5267
I don't know complex analysis though.
Loser66
  • Loser66
You can take i as constant. I use u = 1/t but still get something wrong
Loser66
  • Loser66
let the limits aside, calculate the integrand: du = -1/t^2 dt , hence the integral = \(\dfrac {e^ {(i-1)/t}{i-1} \) , right?
thomas5267
  • thomas5267
Since 0 is within the limits of integral, you might have some problem.
Loser66
  • Loser66
ok, that problem is solving, because it becomes \(lim_{t\rightarrow 0} \int ...\)
thomas5267
  • thomas5267
Maybe rewrite the integral? \[ \int\frac{e^{(i-1)/t}}{t^2}\,dt=\int\frac{e^{i/t}}{e^{1/t}t^2}\,dt=\int\frac{\cos\left(\frac{1}{t}\right)+i\sin\left(\frac{1}{t}\right)}{e^{1/t}t^2} \] Still blows up to infinity though.
thomas5267
  • thomas5267
\[ \int_a^1 \frac{e^{(i-1)/t}}{t^2}\,dt=-\int_\frac{1}{a}^1 e^{(i-1)u}\,du \]
Loser66
  • Loser66
we have e^iu =1 , hence just calculate e^ -u
thomas5267
  • thomas5267
Why e^iu = 1?
anonymous
  • anonymous
e^(i/t)/e^(1/t) let u = 1/t du = -1/t^2
Loser66
  • Loser66
because it is unit circle with radius 1
Loser66
  • Loser66
yes @pgpilot326 , next?
thomas5267
  • thomas5267
No way it is one because 1/a need not equal to 1.
thomas5267
  • thomas5267
It is only part of the unit circle isn't it?
anonymous
  • anonymous
-e^u(i-1)du integrate
anonymous
  • anonymous
you're doing contour over circle with radius of 1 centered on the origin?
thomas5267
  • thomas5267
Isn't the lower limit a and upper limit 1 in the original integral?
anonymous
  • anonymous
can't tell from what was posted, that's why i asked. what is it exactly?
thomas5267
  • thomas5267
No guarantee a is of the form 2pi +1.
thomas5267
  • thomas5267
The original integral should be \[\int_a^1 \dfrac{e^{(i-1)/t}}{t^2}dt\] judging from his incorrect LaTeX. He missed a } after (i-1)/t.
anonymous
  • anonymous
\[-\int\limits_{\frac{ 1 }{a }}^{1}e^{u \left( i-1 \right)}du=\int\limits_{ 1 }^{\frac{1}{a }}e^{u \left( i-1 \right)}du=\int\limits_{ 1 }^{\frac{1}{a }}e^{\left(u i-u\right)}du=\int\limits_{ 1 }^{\frac{1}{a }}\frac{ \left( \cos \left( u \right) \right)-i\sin \left( u \right) }{ e^u }du\]
thomas5267
  • thomas5267
If a>0, then the substitution is continuous and differentiable and allowed. If a<0, we will have some trouble.
anonymous
  • anonymous
let u = -v du = -dv then cos(u)/e^u => -e^v cos(v)dv cos is even function int{e^x cos(x) dx} = 1/2 e^x (sin(x) + cos(x))
anonymous
  • anonymous
do same for sin int{e^x sin(x)} = 1/2 e^x (sin(x) - cos(x))
anonymous
  • anonymous
true, so condition and split up the cases
anonymous
  • anonymous
good luck, gotta go!
thomas5267
  • thomas5267
I am not so sure about splitting up the case since if a=0, 1/a=infinity and still problematic.
anonymous
  • anonymous
limit as a-> 0 from right (1/a, 1) if a < 0, 0 is not a problem... only when a = 0
anonymous
  • anonymous
also, if 0< a < 1, flip the integral if a < 0, not an issue
anonymous
  • anonymous
a is constant
anonymous
  • anonymous
c ya
thomas5267
  • thomas5267
I see, cya.

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