Thread: Laplace transform of piecewise function: Find L {f(t)}, f(t)= 0 ; 0<t<1 t ; t>1

1. Laplace transform of piecewise function: Find L {f(t)}, f(t)= 0 ; 0<t<1 t ; t>1

Find L {f(t)}

f(t)= 0 ; 0<t<1
t ; t>1

My working:
Laplace.jpg
And then I solved for the definite integral. Am I on the right track or have I missed something? Also, is there an easier way to type this on here? It took me close to 45 minutes to be able to type it and upload it to the forums...

2. Originally Posted by promitheus
… is there an easier way to type this on here? It took me close to 45 minutes to be able to type it and upload it …
I can help with this part. You have a couple choices.

(1) Type the expressions directly into your post. Like this (first line):

L{ f(t) } = int[0..infinity] e^(-rt) * f(t) dt

In the Read Before Posting announcement, there is a link titled, "Formatting Math as Text". You can use that reference to learn how to express math statements with a keyboard.

(2) Learn the LaTex coding language, for mathematical formatting. You can google "latex math examples", to look for guides and tutorials.

At this forum, LaTex code must be enclosed within $\text{$$and$}$$ tags.

You may right-click on any LaTex formatting that you see in a post, to view the code (without the tags).

PS: Clicking on the image thumbnail in your post results in a display that's almost too small to read. This is because you placed everything into a single image with lots of blank, white space. The forum software reduces large images. To achieve better results, break large images into two or more smaller images, and crop off unused space around the margins before uploading.

3. Okay, so, the following is what I think is contained in your image, as best as I can read it:

$\displaystyle L \left\{ f(t) \right\} = \int_0^{\infty} e^{-st} f(t) \: dt$

$\displaystyle = \int_0^1 e^{-st} f(t) \: dt + \int_1^{\infty} e^{-st} f(t) \: dt$

$\displaystyle = \int_0^{\infty} e^{-st} 0 \: dt + \int_1^{\infty} e^{-st} \: t \: dt$

$\displaystyle = 0 + \lim_{b \to \infty} \int_1^b e^{-st} \: t \: dt$

$\displaystyle = \lim_{b \to \infty} \int_1^b e^{-st} \: t \: dt$

Let u = t so du = dt

Let $dv=e^{-st} \: dt \text{ so } v=\dfrac{-e^{-st}}{s}$

Integration by parts

$\displaystyle \int_1^b u \: dv = uv - \int_1^b v \: du$

$\displaystyle = t \dfrac{-e^{-st}}{s} - \int_1^b \dfrac{-e^{-st}}{s} \: dt$

$\displaystyle = -t \dfrac{e^{-st}}{s} - \dfrac{-e^{-st}}{s^2}$
Assuming the above is accurate, I agree with your computations thus far, except for one minor sign error in the very last step. The actual indefinite integral should be:

$\displaystyle \int \dfrac{-e^{-st}}{s} \: dt = \dfrac{e^{-st}}{s^2}$

From here, if you follow your next step of "solv[ing] for the definite integral" by evaluating

$\displaystyle = -t \dfrac{e^{-st}}{s} - \dfrac{e^{-st}}{s^2}$

from $t=1 \text{ to } t=\infty$, you'll get the correct answer.

4. I want your Laplace integral to start at zero. Personally, I would think about another variable substitution so that it does.

5. Thanks a lot for all your help. Haven't had to learn math online before, so thanks for the tips on the math typing.

Originally Posted by tkhunny
I want your Laplace integral to start at zero. Personally, I would think about another variable substitution so that it does.
I'm not sure I understand which step you are referring to. Could you please elaborate?

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