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bitrex

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- Thread starter bitrex
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In summary: Thanks for your help.In summary, if you integrate by parts on the function first to break it up, you change the original integral into another integral plus a difference part. You don't want to do this, but if you do the integration by parts and the parts integral is in the table, then you've figured out the original integral.

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bitrex

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Dick

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bitrex

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Dick

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bitrex

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rock.freak667

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Now if L[f(t)]=F(s), then there is a property that says

L[-tf(t)]=F'(s)

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Dick

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bitrex said:integralof the original function, not the original function itself?

You don't want to integrate the function by parts, yes, that would mean you are finding the laplace transform of a different function. That would be foolish. You want to integrate the function times exp(-s*t) by parts. I think that's what the instructions are really trying to tell you. Is that what's confusing?

Last edited:

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bitrex

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Yes, it makes sense now. I had some kind of mental block on what the problem set was asking me to do, heh. So for the [tex]te^{t}[/tex] example I get something like

[tex] \int_0^\infty e^{-st}te^{t}dt = \int_0^\infty e^{-(s-1)t}t dt = \frac{e^{-(s-1)t}}{1 -s}t \bigg|_0^\infty - \frac{1}{1-s} \int_0^\infty e^{-st}e^{t}[/tex]

The second part of the expression is just the Laplace transform of e^t, so we get

[tex]\frac{1}{s^2 - 2s + 1}[/tex]. How's that look?

edit: I've made an error in the above, have to try again

edit: I think the answer is correct now...

[tex] \int_0^\infty e^{-st}te^{t}dt = \int_0^\infty e^{-(s-1)t}t dt = \frac{e^{-(s-1)t}}{1 -s}t \bigg|_0^\infty - \frac{1}{1-s} \int_0^\infty e^{-st}e^{t}[/tex]

The second part of the expression is just the Laplace transform of e^t, so we get

[tex]\frac{1}{s^2 - 2s + 1}[/tex]. How's that look?

edit: I've made an error in the above, have to try again

edit: I think the answer is correct now...

Last edited:

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Dick

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bitrex said:

[tex] \int_0^\infty e^{-st}te^{t}dt = \int_0^\infty e^{-(s-1)t}t dt = \frac{e^{-(s-1)t}}{1 -s}t \bigg|_0^\infty - \frac{1}{1-s} \int_0^\infty e^{-st}e^{t}[/tex]

The second part of the expression is just the Laplace transform of e^t, so we get

[tex]\frac{1}{s^2 - 2s + 1}[/tex]. How's that look?

edit: I've made an error in the above, have to try again

edit: I think the answer is correct now...

That looks just fine.

The Laplace Transform is a mathematical operation that transforms a function of time into a function of complex frequency. It is useful in scientific research because it allows for the analysis and manipulation of complex systems, making it easier to solve differential equations and study dynamic systems.

The Laplace Transform is calculated by integrating the function of time multiplied by the exponential function e^-st, where s is a complex variable. Its key properties include linearity, time-shifting, and convolution, which allow for simplification and manipulation of equations.

No, the function must be well-behaved and have a finite number of discontinuities and bounded growth. Otherwise, the Laplace Transform may not converge or give incorrect results.

The inverse Laplace Transform is the operation that converts a function of complex frequency back into a function of time. It is calculated by using the Bromwich integral, which involves contour integration in the complex plane, or by using a table of known transforms.

The Laplace Transform is commonly used in fields such as engineering, physics, and mathematics, particularly in the study of control systems, circuits, and differential equations. It is also used in signal processing, economics, and other areas of applied mathematics.

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