# Find an integral!

#### evinda

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

#### ZaidAlyafey

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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

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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

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

#### evinda

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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

##### Well-known member
MHB Math Helper
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

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

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

Last edited:

#### evinda

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MHB Site Helper
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

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

#### ZaidAlyafey

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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

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MHB Site Helper
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

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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

##### MHB Seeker
Staff member
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

##### Well-known member
MHB Math Helper
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

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

#### evinda

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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 $$#### Klaas van Aarsen ##### MHB Seeker Staff member Can I also applicate the partial integration rule at the integral$$\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}}$$#### evinda ##### Well-known member MHB Site Helper Using partial integration, let's define:$$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}}$$But,the interval is [a,\infty] .Isn't there a difference? #### Klaas van Aarsen ##### MHB Seeker Staff member But,the interval is [a,\infty] .Isn't there a difference? Oops, I missed the a. But... perhaps you can redo the calculation with an a instead of a 0? Or alternatively, you can use the Laplace table and apply the result for a delayed nth power. #### evinda ##### Well-known member MHB Site Helper Oops, I missed the a. But... perhaps you can redo the calculation with an a instead of a 0? Or alternatively, you can use the Laplace table and apply the result for a delayed nth power. Will it be then,$$I_{n}=I_{0}\frac{n!}{s^{n}}+\frac{e^{-sa}(1-a^{n})}{(1-a)s} $$?Or am I wrong? Last edited: #### evinda ##### Well-known member MHB Site Helper Oops, I missed the a. But... perhaps you can redo the calculation with an a instead of a 0? Or alternatively, you can use the Laplace table and apply the result for a delayed nth power. So...now,I made the calculations and I found that:$$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}$$. But...how can I find this sum?? #### ZaidAlyafey ##### Well-known member MHB Math Helper So...now,I made the calculations and I found that:$$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

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MHB Math Helper
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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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 $$\displaystyle \mathcal{L}(t^n)$$.