Quick Laplace Transform theory question

• bitrex
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.
bitrex
I have some problems where I am to take the Laplace transform of some functions, for example $$te^{xt}$$ not by using the integral definition but instead by using a table of transforms and the fact that the Laplace transform is a linear operator. However, the instructions say that "A preliminary integration by parts may be necessary." If I use an integration by parts on the function first though to break it up, haven't I changed the function to something other than what I am supposed to be taking the Laplace transform of? Thanks for any advice.

If you do an integration by parts you change the original integral into another integral plus a difference part. Sure, you have to keep track of the difference part if you expect to get the laplace transform of the original function.

I'm afraid I'm still not entirely clear on what to do - apparently the purpose of the exercise is to calculate the Laplace transform just from the tables without using the integral definition at all, so I'm not completely clear on why I'm integrating something anyway. Essentially the instructions say "Don't use the integral definition of the Laplace transform, but you may have to integrate" so I'm a bit confused.

If you do the integration by parts, and the parts integral is in the table, then you've figured out the original integral. Why don't you just try an example?

Ok, for example, one of the functions that I can't apply linearity to directly is the Laplace transform of $$te^{t}$$. If I integrate that by parts I get $$te^{t} -\int \!e^{t}$$. That doesn't seem to simplify the problem as I'm left with something worse than what I started with. Also, even if I could use the tables to find the Laplace transform of this new function, wouldn't I be finding the Laplace transform of the integral of the original function, not the original function itself?

Well you know that L[ekt]=1/(s-k) for s>0.

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

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

bitrex said:
Ok, for example, one of the functions that I can't apply linearity to directly is the Laplace transform of $$te^{t}$$. If I integrate that by parts I get $$te^{t} -\int \!e^{t}$$. That doesn't seem to simplify the problem as I'm left with something worse than what I started with. Also, even if I could use the tables to find the Laplace transform of this new function, wouldn't I be finding the Laplace transform of the integral of 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?

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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 $$te^{t}$$ example I get something like

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

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

$$\frac{1}{s^2 - 2s + 1}$$. 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:
bitrex said:
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 $$te^{t}$$ example I get something like

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

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

$$\frac{1}{s^2 - 2s + 1}$$. 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.

1. What is the Laplace Transform and why is it useful in scientific research?

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.

2. How is the Laplace Transform calculated and what are its key properties?

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.

3. Can the Laplace Transform be applied to any function?

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.

4. What is the inverse Laplace Transform and how is it calculated?

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.

5. In what fields of science is the Laplace Transform commonly used?

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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