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- #1

- Apr 13, 2013

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Hey!!!

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

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

- Thread starter evinda
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- Thread starter
- #1

- Apr 13, 2013

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Hey!!!

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

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

- Jan 17, 2013

- 1,667

- Thread starter
- #3

- Apr 13, 2013

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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?

- Mar 22, 2013

- 573

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- #5

- Apr 13, 2013

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

How can I use this??

- Jan 17, 2013

- 1,667

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 integralBut..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?

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

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

- Mar 22, 2013

- 573

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

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

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

Last edited:

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- #8

- Apr 13, 2013

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

- Mar 22, 2013

- 573

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

- Jan 17, 2013

- 1,667

Well, if you are findingSo,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?

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

But for

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

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- #11

- Apr 13, 2013

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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??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

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

- Mar 22, 2013

- 573

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- #13

- Mar 5, 2012

- 8,887

That is correct, although it is nice to have a normal expression for it.So,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?

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

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:

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

- Jan 17, 2013

- 1,667

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 followingI 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??

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

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

- Mar 22, 2013

- 573

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- #16

- Apr 13, 2013

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

$$\mathcal L\{t^{n}\} = \frac{n!}{s^{n+1}}$$

AsZaidAlyafeysuggested, 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:

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

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- #17

- Mar 5, 2012

- 8,887

Using partial integration, let's define:Can I also applicate the partial integration rule at the integral $$\int_a^\infty e^{-st}t^n dt $$

$$I_n = \int_0^\infty e^{-st}t^n dt$$

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:

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

- Thread starter
- #18

- Apr 13, 2013

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

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

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- #19

- Mar 5, 2012

- 8,887

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

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.

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- #20

- Apr 13, 2013

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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?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.

Last edited:

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- #21

- Apr 13, 2013

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So...now,I made the calculations and I found that: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.

$$I_{n}=e^{-sa}\Sigma_{k=1}^{n}\frac{a^{n+1-k}n!}{s^{2k-1}(n-k+1)!}+\frac{n!}{s^{2n}}I_{0} \text{ ,where } I_{0}=\frac{e^{-sa}}{s}$$ .

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

- Jan 17, 2013

- 1,667

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

- Jan 17, 2013

- 1,667

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.

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- #24

- Apr 13, 2013

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

- Jan 17, 2013

- 1,667

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 \(\displaystyle \mathcal{L}(t^n)\).I have found the result,using partial integration.Do you think that it is wrong?