Find Integral: $\int_{a}^{\infty}e^{-st}t^{n}dt$

[evinda]

Hey! :)
Could you explain me how I can find the integral $\int_{a}^{\infty}e^{-st}t^{n}dt$?

 

[alyafey22]

You need to eliminate $t^n$ by multiple differentiation using integration by parts. Start by integrating $e^{-st}$ then differentiate $t^n$ and so one until $t^n$ vanishes.

 

[evinda]

You need to eliminate $t^n$ by multiple differentiation using integration by parts. Start by integrating $e^{-st}$ then differentiate $t^n$ and so one until $t^n$ vanishes.

But..isn't there also an other way?Because the value of n is not given.So,how can I know how many times I have to make integration by parts?

 

[mathbalarka]

There surely are other ways. That thing there is related to incomplete gamma function, upto some constant factor.

 

[evinda]

There surely are other ways. That thing there is related to incomplete gamma function, upto some constant factor.

It is like that,right? $\Gamma(p+1)=\int_{0}^{\infty}e^{-x}x^{p}dx$
How can I use this??

 

[alyafey22]

But..isn't there also an other way?Because the value of n is not given.So,how can I know how many times I have to make integration by parts?

Yes , there are other ways. As Balarka said using Incomplete Gamma function. Since, you are most problably interested in elementary techniques this is the most standard way. Since n is a positive integer (it should be due to convergence issues), it decreases by one each time IBP is employed. You can then proceed by induction. Actually for the complete integral

$$\int^{\infty}_0 e^{-st}t^{n} \, dt=\frac{n!}{s^{n+1}}$$

The proof of this could be done using IBP or the Gamma function.

 

[mathbalarka]

That is gamma function you are using there. See this. For integer arguments this is really equal to

$$\frac{n!}{s^{n+1}}$$

 

[evinda]

Yes , there are other ways. As Balarka said using Incomplete Gamma function. Since, you are most problably interested in elementary techniques this is the most standard way. Since n is a positive integer (it should be due to convergence issues), it decreases by one each time IBP is employed. You can then proceed by induction. Actually for the complete integral

$$\int^{\infty}_0 e^{-st}t^{n} \, dt=\frac{n!}{s^{n+1}}$$

The proof of this could be done using IBP or the Gamma function.

So,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?

 

[mathbalarka]

Yes, you can denote is as a Laplace transform instead, that is correct.

 

[alyafey22]

So,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?

Well, if you are finding

$$\mathcal{L}(t^n) = \int^{\infty}_0 e^{-st}t^n \, dt= \frac{\Gamma(n+1)}{s^{n+1}}$$

But for

$$\int^{\infty}_a e^{-st}t^n\, dt = \frac{\Gamma(n+1,a)}{s^{n+1}}$$

 

[evinda]

Well, if you are finding

$$\mathcal{L}(t^n) = \int^{\infty}_0 e^{-st}t^n \, dt= \frac{\Gamma(n+1)}{s^{n+1}}$$

But for

$$\int^{\infty}_a e^{-st}t^n\, dt = \frac{\Gamma(n+1,a)}{s^{n+1}}$$

I haven't get taught the incomplete gamma function.Couldn't I just use the gamma function,or do I have to use the incomplete gamma function??

 

[mathbalarka]

I remember having read somewhere about inexpressibility of incomplete gamma in terms of gamma using complicated transcendence theory but I cannot recall it now, so don't rely on this post much. Z is the expert here.

 

[I like Serena]

So,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?

That is correct, although it is nice to have a normal expression for it.
The standard (engineering) way to find Laplace transforms, is to look it up in a Table of Laplace transforms.
Here you can find that:
$$\mathcal L\{t^{n}\} = \frac{n!}{s^{n+1}}$$

As ZaidAlyafey[/color] suggested, the standard approach to calculate it yourself would be repeated application of the partial integration rule.

Or alternatively:
\begin{aligned}\frac{d^n}{ds^n} \int_0^\infty e^{-st}dt
&= \int_0^\infty \frac{d^n}{ds^n} e^{-st}dt \\
&= (-1)^n \int_0^\infty e^{-st}t^n dt \end{aligned}

Since we also have:
\begin{aligned} \frac{d^n}{ds^n} \int_0^\infty e^{-st}dt
&= \frac{d^n}{ds^n}\Big( \frac {-e^{-st}}{s}\Bigg|_0^\infty \Big) \\
&= \frac{d^n}{ds^n}\Big( \frac {1}{s} \Big) \\
&= (-1)^n \frac{n!}{s^{n+1}} \end{aligned}

It follows that:
$$\int_0^\infty e^{-st}t^n dt = \frac{n!}{s^{n+1}}$$

 

[alyafey22]

I haven't get taught the incomplete gamma function.Couldn't I just use the gamma function,or do I have to use the incomplete gamma function??

Incomplete Gamma function is difficult to deal with. as I said the easiest way is IBP. I am busy at the moment to derive a formula but for easiness try the following

$$\int^{\infty}_a e^{-t} t^n \, dt$$

If you cannot derive a formula, I can do it later.

 

[mathbalarka]

(Not really on-topic but : Shh! Don't disturb Z, he is trying to get his INC4 integration result right :p)

 

[evinda]

That is correct, although it is nice to have a normal expression for it.
The standard (engineering) way to find Laplace transforms, is to look it up in a Table of Laplace transforms.
Here you can find that:
$$\mathcal L\{t^{n}\} = \frac{n!}{s^{n+1}}$$

As ZaidAlyafey[/color] suggested, the standard approach to calculate it yourself would be repeated application of the partial integration rule.

Or alternatively:
\begin{aligned}\frac{d^n}{ds^n} \int_0^\infty e^{-st}dt
&= \int_0^\infty \frac{d^n}{ds^n} e^{-st}dt \\
&= (-1)^n \int_0^\infty e^{-st}t^n dt \end{aligned} $$ ?

Since we also have:
\begin{aligned} \frac{d^n}{ds^n} \int_0^\infty e^{-st}dt
&= \frac{d^n}{ds^n}\Big( \frac {-e^{-st}}{s} dt \Bigg|_0^\infty \Big) \\
&= \frac{d^n}{ds^n}\Big( \frac {1}{s} \Big) \\
&= (-1)^n \frac{n!}{s^{n+1}} \end{aligned}

It follows that:
$$\int_0^\infty e^{-st}t^n dt = \frac{n!}{s^{n+1}}$$

Can I also applicate the partial integration rule at the integral $$\int_a^\infty e^{-st}t^n dt $$

 

[I like Serena]

Can I also applicate the partial integration rule at the integral $$\int_a^\infty e^{-st}t^n dt $$

Using partial integration, let's define:
$$I_n = \int_0^\infty e^{-st}t^n dt$$

Then we have with partial integration that:
\begin{array}{ccccc}
I_n &=& \int_0^\infty t^n d\Big(\frac{-e^{-st}}{s}\Big) \\
&=& t^n\frac{-e^{-st}}{s}\Bigg|_0^\infty &+& \int_0^\infty \frac{e^{-st}}{s}d(t^n) \\
&=& 0 &+& \frac n s \int_0^\infty e^{-st} t^{n-1}dt \\
&=& \frac n s I_{n-1} \end{array}

It follows that:
$$I_{n-1} = \frac {n-1} s I_{n-2}$$
$$I_{n-2} = \frac {n-2} s I_{n-3}$$
$$...$$

Substituting, we get that:
$$I_n = \underbrace{\frac n s \cdot \frac {n-1} s \cdot \frac {n-2} s \cdot ... \cdot \frac 1 s}_{n\text{ factors}} \cdot I_0 = \frac{n!}{s^n} \cdot I_0$$

For $I_0$ we have that:
$$I_0 = \int_0^\infty e^{-st} dt = \frac{-e^{-st}}{s}\Bigg|_0^\infty = \frac 1 s$$

So:
$$I_n = \frac {n!}{s^{s+1}}$$

 

[evinda]

Using partial integration, let's define:
$$I_n = \int_0^\infty e^{-st}t^n dt$$

Then we have with partial integration that:
\begin{array}{ccccc}
I_n &=& \int_0^\infty t^n d\Big(\frac{-e^{-st}}{s}\Big) \\
&=& t^n\frac{-e^{-st}}{s}\Bigg|_0^\infty &+& \int_0^\infty \frac{e^{-st}}{s}d(t^n) \\
&=& 0 &+& \frac n s \int_0^\infty e^{-st} t^{n-1}dt \\
&=& \frac n s I_{n-1} \end{array}

It follows that:
$$I_{n-1} = \frac {n-1} s I_{n-2}$$
$$I_{n-2} = \frac {n-2} s I_{n-3}$$
$$...$$

Substituting, we get that:
$$I_n = \underbrace{\frac n s \cdot \frac {n-1} s \cdot \frac {n-2} s \cdot ... \cdot \frac 1 s}_{n\text{ factors}} \cdot I_0 = \frac{n!}{s^n} \cdot I_0$$

For $I_0$ we have that:
$$I_0 = \int_0^\infty e^{-st} dt = \frac{-e^{-st}}{s}\Bigg|_0^\infty = \frac 1 s$$

So:
$$I_n = \frac {n!}{s^{s+1}}$$

But,the interval is $[a,\infty]$ .Isn't there a difference?

 

[I like Serena]

But,the interval is $[a,\infty]$ .Isn't there a difference?

Oops, I missed the $a$.
But... perhaps you can redo the calculation with an $a$ instead of a $0$?

Or alternatively, you can use the Laplace table and apply the result for a delayed nth power.

 

[evinda]

Oops, I missed the $a$.
But... perhaps you can redo the calculation with an $a$ instead of a $0$?

Or alternatively, you can use the Laplace table and apply the result for a delayed nth power.

Will it be then,$$I_{n}=I_{0}\frac{n!}{s^{n}}+\frac{e^{-sa}(1-a^{n})}{(1-a)s} $$?Or am I wrong?

 

[evinda]

Oops, I missed the $a$.
But... perhaps you can redo the calculation with an $a$ instead of a $0$?

Or alternatively, you can use the Laplace table and apply the result for a delayed nth power.

So...now,I made the calculations and I found that:
$$I_{n}=e^{-sa}\Sigma_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{2k-1}(n-k+1)!}+\frac{n!}{s^{2n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$ .

But...how can I find this sum??

 

[alyafey22]

So...now,I made the calculations and I found that:
$$I_{n}=e^{-sa}\Sigma_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{2k-1}(n-k+1)!}+\frac{n!}{s^{2n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$ .

But...how can I find this sum??

Well, you are expected to have a finite sum. You don't need to evaluate it. Verify the result for chosen values of $(n,s,a)$.

 

[alyafey22]

Actually since we are working on Incomplete gamma function , we will not get a nice looking expression as for the gamma function. The gamma function goes like $$\frac{1}{n(n-1)(n-2) \cdots 1}=\frac{1}{n!}$$

But Incomplete gamma function as the name expresses doesn't work as nice. Actually we expect something close to the $$ \sum {n \choose k } $$. But I haven't yet verified your expression.

 

[evinda]

I have found the result,using partial integration.Do you think that it is wrong?

 

[alyafey22]

I have found the result,using partial integration.Do you think that it is wrong?

Try it for some values or post the full solution and I can check it for you. Remember that for $a \to 0 $ the result should converge to $$\mathcal{L}(t^n)$$.

 

[evinda]

Is there not a way I could calculate the sum I found??

 

[I like Serena]

So...now,I made the calculations and I found that:
$$I_{n}=e^{-sa}\Sigma_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{2k-1}(n-k+1)!}+\frac{n!}{s^{2n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$ .

But...how can I find this sum??

Why should it converge to $$\mathcal{L}(t^n)$$ for $a \to 0 $ ?

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Is there not a way I could calculate the sum I found??

I get:
$$I_{n}=e^{-sa}\sum_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{k+1}(n-k+1)!}+\frac{n!}{s^{n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$
To be honest, I don't know how to calculate the sum.

Now, to verify part of it.
We already know that with $a=0$ we should get $\mathcal L\{t^n\} = \frac{n!}{s^{n+1}}$.
For $a=0$ the summation in your expression is zero, leaving only the last part.
This is where you can see that your $s^{2n}$ should really be $s^n$.Alternatively, you can write that:
$$I_n = \mathcal L\{t^n u(t-a)\} = e^{-sa} \mathcal L\{(t+a)^n\}$$
where $u(t)$ is the unit step function (0 for a negative argument, 1 for a positive argument).

 

[evinda]

I get:
$$I_{n}=e^{-sa}\Sigma_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{k+1}(n-k+1)!}+\frac{n!}{s^{n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$
To be honest, I don't know how to calculate the sum.

Now, to verify it.
We already know that with $a=0$ we should get $\mathcal L(t^n) = \frac{n!}{s^{n+1}}$.
For $a=0$ the summation in your expression is zero, leaving only the last part.
This is where you can see that your $s^{2n}$ should really be $s^n$.

I found that $$I_{n}=\frac{e^{-sa}a^{n}}{s}+\frac{nI_{n-1}}{s^{2}}$$
$$\text{so, }I_{n-1}=\frac{e^{-sa}a^{n-1}}{s}+\frac{(n-1)I_{n-2}}{s^{2}}$$ ans so on.What have I done wrong??

 

[I like Serena]

I found that $$I_{n}=\frac{e^{-sa}a^{n}}{s}+\frac{nI_{n-1}}{s^{2}}$$
$$\text{so, }I_{n-1}=\frac{e^{-sa}a^{n-1}}{s}+\frac{(n-1)I_{n-2}}{s^{2}}$$ ans so on.What have I done wrong??

I get:
$$I_{n}=\frac{e^{-sa}a^{n}}{s}+\frac{nI_{n-1}}{s}$$

 

[evinda]

I get:
$$I_{n}=\frac{e^{-sa}a^{n}}{s}+\frac{nI_{n-1}}{s}$$

Oh sorry!I accidentally wrote at one equation twice the $s$..
Now,I get $$I_{n}=e^{-sa}\Sigma_{k=1}^{n} \frac{a^{n+1-k}}{s^{k}}\frac{n!}{(n-k+1)!}+\frac{n!I_{0}}{s^{n}}$$ .
What have I done wrong now??

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We already know that with $a=0$ we should get $\mathcal L\{t^n\} = \frac{n!}{s^{n+1}}$.

And also...why do we know that with $a=0$ we should get $\mathcal L\{t^n\} = \frac{n!}{s^{n+1}}$ ?