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## Differential equations

### Course: Differential equations > Unit 3

Lesson 2: Properties of the Laplace transform- Laplace as linear operator and Laplace of derivatives
- Laplace transform of cos t and polynomials
- "Shifting" transform by multiplying function by exponential
- Laplace transform of t: L{t}
- Laplace transform of t^n: L{t^n}
- Laplace transform of the unit step function
- Inverse Laplace examples
- Dirac delta function
- Laplace transform of the dirac delta function

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# Laplace transform of t^n: L{t^n}

Laplace Transform of t^n: L{t^n}. Created by Sal Khan.

## Want to join the conversation?

- Haven't you done this in an earlier video? Maybe not as rigorous...(15 votes)
- Yes, but last time he used the formula L(f)=(1/s)[L(f')+f(0)] to solve for it.(12 votes)

- At 3.03 Sal put the boundary condition of 0 to inf in blue box. But I don't find any integration of that term in formula, so why these conditions??(4 votes)
- that's the definition of integration by parts for a definite integral.(2 votes)

- L(t^n) if n is not=+ve intiger??(4 votes)
- In the first term of the equation (around the4:30min mark), Sal assumes e^(-st) goes to zero faster than t^(n) getting bigger when t go to infinity, isn't that also depends on the value of n? If n is huge, t^(n) might grow faster than e^(-st)?? If that is the case, this term won't go to zero?(1 vote)
- No, it doesn't depend on n. All exponential functions move faster than all polynomial functions. If n is huge, it may take a long time for the e⁻ˢᵗ to overcome the tⁿ, but it will happen eventually, (assuming of course that s > 0).

This can be easily proven by checking the growth rates of the product. Every time t increases by 1, the e⁻ˢᵗ multiplies by e⁻ˢ, a constant less than 1, but the tⁿ term multiplies by (1 + 1/t)ⁿ, which approaches 1 as t gets large, regardless of n. So, as t gets large, the whole function is multiplied by a number less than 1 for each incremental increase in t. Therefore, as t approaches infinity, the function approaches 0, regardless of n.(3 votes)

- can we get a video incorporating the gamma function and laplace transforms? Thanks!(2 votes)
- Why does s have to be larger than 0? I can see why it cant be 0, but why cant it be smaller than 0?(1 vote)
- If s is smaller than 0, ∫e^(-st)*f(t) dt from 0 to ∞ will diverge. For example, if you use integration by parts, you would have to evaluate e^(-st)*f'(t) in the first term from 0 to ∞. Since s positive, the limit of e^(-st) as t approaches ∞ would be 0 (you can show this by graphing e^x and looking at where x is considerably smaller than 0). However, if s were negative, the limit of e^(-st) as t approaches ∞ would diverge (you can show this by graphing e^x and looking at where x is considerably larger than 0).(1 vote)

- This is cool. However, what if an entire function has an exponent? How do we take the Laplace Transform of an arbitrary function squared? Does the nature of the function need to be known? L {f(x) ^ 2} = ?(1 vote)
- I'm having difficulties with an assignment, finding the laplace transform of (t^(t^2)), any form of help will be highly appreciated(1 vote)
- Why do we say n>0 when it technically works when n=0?(1 vote)
- is laplace transformation and gamma function similar?(1 vote)

## Video transcript

In the last video, I showed the
Laplace transform of t, or we could view that as t the
first power, is equal to 1/s squared, if we assume that
s is greater than 0. In this video, we're going to
see if we can generalize this by trying to figure out the
Laplace transform of t to the n, where n is any integer power
greater than 0, so n is any positive integer
greater than 0. So let's try it out. So we know from our definition
of the Laplace transform that the Laplace transform of t to
the n is equal to the integral from 0 to infinity of our
function-- well, let me write t to the n-- times, and this is
just the definition of the transform, e to the
minus st, dt. And similar to when we figured
out this Laplace transform, your intuition might be that,
hey, we should use integration by parts, and I showed
it in the last video. I always forget it, but I just
recorded it, so I do happen to remember it. So the integration by parts
just tells us that the integral of uv prime is equal to
uv minus the integral of-- I view this as kind of the
swap-- so u prime v. So this is just our integration
by parts formula. If you ever forget it, you can
derive it in about 30 seconds from the product rule. And I did it in the last video
because I hadn't used it for awhile, so I had
to rederive it. So let's apply it here. So what do we want to
make our v prime? It's always good to use the
exponential function, because that's easy to take the
antiderivative of. So this is our v prime, in which
case our v is just the antiderivative of that. So it's e to the minus
st over minus s. If we take the derivative of
this, minus s divided by minus s cancels out, and you
just get that. And then if we make our u-- let
me pick a good color here. If we make this equal to our
u, what's our u prime? u prime is just going to be n
times t to the n minus 1. Fair enough. So let's apply the integration
by parts. So this is going to
be equal to uv. u, I'll use this t to the n, so
u is t to the n, that's our u, times v, which is e-- let me
write this down-- so it's minus-- there's a minus sign
there, so we put the minus. Let me do it in that color. I'm just rewriting this.
e to the minus st/s. So that's the uv term
right there. Let me make that clear. And let me pick a
good color here. So this term right here is
this term right here. And, of course, I'm going to
have to evaluate this from 0 to infinity, so let me write
that: 0 to infinity. I could put a little bracket
there or something, but you know we're going to have
to evaluate that. And then from that, we're
going to have to subtract the integral. And let me not forget
our boundaries. 0 to infinity of u prime is n
times t to the n minus 1-- that's our u prime-- times v
times minus-- so let me put this minus out here-- so minus
e to the minus st/s. And then all of that. Of course, we have our dt, and
you have a minus minus. These things become pluses. Let's see if we can simplify
this a little bit. So we get our Laplace transform
of t to the n is equal to this evaluated at
infinity and evaluated at 0. So when you've evaluate-- what's
the limit of this as t approaches infinity? As t approaches infinity, this
term, you might say, oh, this becomes really big. And I went over this
in the last video. But this term overpowers it,
because you're going to have e to the minus infinity, if
we assume that s is greater than 0. So if s is greater than 0, this
term is going to win out and go to 0 much faster
than this term is going to go to infinity. So when you evaluate it at
infinity, when you evaluate this at infinity, you're
going to get 0. And then you're going to
subtract this evaluated at 0. This evaluated at 0, when it's
evaluated at 0 is just minus 0 to the n times e to the minus
s times 0/s Well, this becomes 0 as well. So this whole term evaluated
from 0 to infinity is all 0, which is a nice, convenient
thing for us. And then we're going to have
this next term right there. So let's take out the constant
terms. This n and this s are constant. They are constant with
respect to t. So you have plus n/s times the
integral from 0 to infinity of t to the n minus 1 times
e to the minus st, dt. Now, this should look reasonably
familiar to you. What's the definition of
the Laplace transform? The Laplace transform of any
function is equal to the integral from 0 to infinity of
that function times e to the minus st, dt. Well, when we have an e to the
minus st, dt, we're taking the integral from 0 to infinity,
so this whole integral is equal to the Laplace transform
of this, of t to the n minus 1. So just that easily, because
this term went to 0, we've simplified things. We get the Laplace transform
of t to the n is equal to-- this is all 0-- it's equal to
n/s-- that's right there-- times this integral right here,
which we just figured out was the Laplace transform
of t to the n minus 1. Well, this is a nice,
neat simplification. We can now figure out the
Laplace transform of a higher power in terms of the one power
lower that, but it still doesn't give me a generalized
formula. So let's see if we can use this
with this information to get a generalized formula. So the Laplace transform of just
t-- so let me write that down; I wrote that at the
beginning of the problem. We get the Laplace transform, I
could write this as t to the 1, which is just t, is equal
to 1/s squared, where s is greater than 0. Now, what happens if
we take the Laplace transform of t squared? Well, we can just use this
formula up here. The Laplace transform of t
squared is equal to 2/s times the Laplace transform of t,
of just t to the 1, right? 2 minus 1. So times the Laplace transform
of t to the 1. Well, t, we know what that is. This is equal to 2/s times this,
times 1/s squared, which is equal to 2/s to the third. Interesting. Let's see if we can
do another one. What is-- I'll do it in the
dark blue-- the Laplace transform of t to the third? Well, we just use this
formula up here. It's n/s. In this case, n is 3. So it's 3/s times the Laplace
transform of t to the n minus 1, so t squared. We know what the Laplace
transform of this one was. This is just this right there. So it's equal to 3/s
times this thing. And I'm going to actually write
it this way, because I think it's interesting. So I'll write the numerator. Times 2 times 1/s over s
squared, which is-- we could write it as 3 factorial
over-- what is this? s to the fourth power. Let's do another one. And I think you already are
getting the idea of what's going on. The Laplace transform of t to
the fourth power is what? It's equal to 4/s times the
Laplace transform of t the third power. And that's just 4/s
times this. So it's 4/s times 3 factorial
over s to the fourth. So now 4 times 3 factorial,
that's just 4 factorial over s to the fifth. And so you can just get
the general principle. And we can prove this
by induction. It's almost trivial based on
what we've already done. That the Laplace transform of
t to the n is equal to n factorial over s to
the n plus 1. We proved it directly for this
base case right here. This is 1 factorial over
s to the 1 plus 1. And then if we know it's true
for this, we know it's going to be true for the
next increment. So induction proof is almost
obvious, but you can even see it based on this. If you have to figure out the
Laplace transform of t to the tenth, you could just keep
doing this over and over again, but I think you see the
pattern pretty clearly. So anyway, I thought that was
a neat problem in and of itself, outside of the fact,
it'll be useful when we figure out inverse and Laplace
transforms. But this is a pretty neat result. The Laplace transform of t to
the n, where n is some integer greater than 0 is equal to n
factorial over s to the n plus 1, where s is also
greater than 0. That was an assumption we had to
make early on when we took our limits as t approaches
infinity. Anyway, hopefully you
found that useful.