Find an integral!

[evinda]

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

 

[ZaidAlyafey]

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??

 

[ZaidAlyafey]

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

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

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

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

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.

 

[ZaidAlyafey]

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

Well, if you are finding

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

But for

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

 

[evinda]

Well, if you are finding

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

But for

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

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.

 

[Klaas van Aarsen]

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

 

[ZaidAlyafey]

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

\(\displaystyle \int^{\infty}_a e^{-t} t^n \, dt\)

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 )

 

[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 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 $$\int_a^\infty e^{-st}t^n dt $$

 

[Klaas van Aarsen]

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?

 

[Klaas van Aarsen]

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??

 

[ZaidAlyafey]

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 $(n,s,a)$.

 

[ZaidAlyafey]

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 \(\displaystyle \frac{1}{n(n-1)(n-2) \cdots 1}=\frac{1}{n!}\)

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

 

[evinda]

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

 

[ZaidAlyafey]

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 \(\displaystyle \mathcal{L}(t^n)\).