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- 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 $\int_{a}^{\infty}e^{-st}t^{n}dt$?
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?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.
It is like that,right? $\Gamma(p+1)=\int_{0}^{\infty}e^{-x}x^{p}dx$There surely are other ways. That thing there is related to incomplete gamma function, upto some constant factor.
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?![]()
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.
Well, if you are findingSo,I can just say that the integral is equal to $L\{t^{n}\}$ ,right?
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
\(\displaystyle \int^{\infty}_a e^{-st}t^n\, dt = \frac{\Gamma(n+1,a)}{s^{n+1}}\)
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?
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??![]()
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:
$$\mathcal L\{t^{n}\} = \frac{n!}{s^{n+1}}$$
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}}$$
Using partial integration, let's define:Can I also applicate the partial integration rule at the integral $$\int_a^\infty e^{-st}t^n dt $$
But,the interval is $[a,\infty]$ .Isn't there a difference?Using partial integration, let's define:
$$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}}$$
Oops, I missed the $a$.But,the interval is $[a,\infty]$ .Isn't there a difference?
Will it be then,$$I_{n}=I_{0}\frac{n!}{s^{n}}+\frac{e^{-sa}(1-a^{n})}{(1-a)s} $$?Or am I wrong?Oops, I missed the $a$.
But... perhaps you can redo the calculation with an $a$ instead of a $0$?
Or alternatively, you can use the Laplace table and apply the result for a delayed nth power.
So...now,I made the calculations and I found that:Oops, I missed the $a$.
But... perhaps you can redo the calculation with an $a$ instead of a $0$?
Or alternatively, you can use the Laplace table and apply the result for a delayed nth power.
Well, you are expected to have a finite sum. You don't need to evaluate it. Verify the result for chosen values of $(n,s,a)$.So...now,I made the calculations and I found that:
$$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??![]()
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?![]()