How do I solve the Laplace transform of te^t without using a table?

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To solve the Laplace transform of f(t) = te^t without using a table, integration by parts is employed multiple times. The integral is manipulated to ∫ te^(t(1-s)) dt, leading to an expression that simplifies after integration. The evaluation from 1 to infinity indicates that, assuming s > 1, the term te^[t(1-s)] approaches 0 as t approaches infinity. The region of convergence for this Laplace transform is to the right of the line s = 1 in the complex plane. The discussion concludes with the realization that the initial approach was correct, and the solution can be derived from these steps.
Pi Face
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Homework Statement



f(t)=te^t, find laplace

Homework Equations





The Attempt at a Solution



I started doing integration by parts and after doing it three times I wasn't sure if I was going in the right direction/making any progress. I'm not supposed to use a table to solve this (I have to do the integral out) so could anyone give me a hint as to how to start?

I can manipulate the original equation to
(0 to inf for all integrals)
∫ te^(t(1-s)) dt and then i set u=t, du=dt, dv=e^(t(1-s)), v=e^(t(1-s))/(1-s)
and then I end up with a longer expression and I need to integrate by parts again. is this is right direction?
 
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It's the right direction. After all, your t-factor in the integrand goes away, so you just have to integrate e^(t*something).
 
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

EDIT: nevermind, I had a brainfart, I got the answer
 
Last edited:
Pi Face said:
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

Laplace transform generally has a region of convergence to the right of some axis parallel to the y-axis on the complex s plane, in this case this axis happens to be x=1.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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