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Find an integral!

evinda

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Apr 13, 2013
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Hey!!! :)
Could you explain me how I can find the integral $\int_{a}^{\infty}e^{-st}t^{n}dt$?
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
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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Apr 13, 2013
3,723
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? :confused:
 

mathbalarka

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Mar 22, 2013
573
There surely are other ways. That thing there is related to incomplete gamma function, upto some constant factor.
 

evinda

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Apr 13, 2013
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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
Jan 17, 2013
1,667
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? :confused:
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

Well-known member
MHB Math Helper
Mar 22, 2013
573
That is gamma function you are using there. See this. For integer arguments this is really equal to

$$\frac{n!}{s^{n+1}}$$
 
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evinda

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MHB Site Helper
Apr 13, 2013
3,723
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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MHB Math Helper
Mar 22, 2013
573
Yes, you can denote is as a Laplace transform instead, that is correct.
 

ZaidAlyafey

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Jan 17, 2013
1,667
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
Apr 13, 2013
3,723
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?? :confused:
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
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
Mar 5, 2012
8,887
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
Jan 17, 2013
1,667
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?? :confused:
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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MHB Math Helper
Mar 22, 2013
573
(Not really on-topic but : Shh! Don't disturb Z, he is trying to get his INC4 integration result right :p)
 

evinda

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MHB Site Helper
Apr 13, 2013
3,723
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
Mar 5, 2012
8,887
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

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MHB Site Helper
Apr 13, 2013
3,723
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
Mar 5, 2012
8,887
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

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MHB Site Helper
Apr 13, 2013
3,723
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? :confused:
 
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evinda

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Apr 13, 2013
3,723
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?? :eek:
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
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?? :eek:
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
Jan 17, 2013
1,667
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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Apr 13, 2013
3,723
I have found the result,using partial integration.Do you think that it is wrong? :confused:
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
I have found the result,using partial integration.Do you think that it is wrong? :confused:
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 \(\displaystyle \mathcal{L}(t^n)\).