MHB Find Integral: $\int_{a}^{\infty}e^{-st}t^{n}dt$

[I like Serena]

Oh sorry!I accidentally wrote at one equation twice the $s$..
Now,I get $$I_{n}=e^{-sa}\Sigma_{k=1}^{n} \frac{a^{n+1-k}}{s^{k}}\frac{n!}{(n-k+1)!}+\frac{n!I_{0}}{s^{n}}$$ .
What have I done wrong now??

In retrospect I believe that is correct.
I had summed with different indices and I made a mistake when converting those.

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And also...why do we know that with $a=0$ we should get $\mathcal L\{t^n\} = \frac{n!}{s^{n+1}}$ ?

Applying the definition of the Laplace transform we have:
$$\mathcal L\{t^n\} = \int_0^\infty e^{-st}t^n dt$$
This is your integral with $a=0$.

Additionally, in post http://mathhelpboards.com/differential-equations-17/find-integral-8382-2.html#post38771 and post http://mathhelpboards.com/differential-equations-17/find-integral-8382-2.html#post38777, I showed in different ways that it is equal to $$\frac{n!}{s^{n+1}}$$.

 

[evinda]

In retrospect I believe that is correct.

Great!

Applying the definition of the Laplace transform we have:
$$\mathcal L\{t^n\} = \int_0^\infty e^{-st}t^n dt$$
This is your integral with $a=0$.

Additionally, in post http://mathhelpboards.com/differential-equations-17/find-integral-8382-2.html#post38771 and post http://mathhelpboards.com/differential-equations-17/find-integral-8382-2.html#post38777, I showed in different ways that it is equal to $$\frac{n!}{s^{n+1}}$$.

oh yes!Right!And to explain why it is right,can I just say that it is verified for $a=0$ ?

 

[I like Serena]

Great!

oh yes!Right!And to explain why it is right,can I just say that it is verified for $a=0$ ?

To be honest, that is only very little verification, since it doesn't touch the sum-part of the expression.
For a proper verification, I think you should also verify it for say $a=1$ and $n=1$ and also for $a=2$ and $n=3$.
Easy enough to do with W|A.

 

[evinda]

To be honest, that is only very little verification, since it doesn't touch the sum-part of the expression.
For a proper verification, I think you should also verify it for say $a=1$ and $n=1$ and also for $a=2$ and $n=3$.
Easy enough to do with W|A.

I checked it for the values $a=1$ and $n=1$ and $a=2$ and $n=3$,and the results were the same with the Laplace transform! :) So,do I have to write these two examples,for verification?

 

[I like Serena]

I checked it for the values $a=1$ and $n=1$ and $a=2$ and $n=3$,and the results were the same with the Laplace transform! :)

Good! ;)

So,do I have to write these two examples,for verification?

No need. It suffices that you are sure that your formula is correct.
You should write down the steps how you got to the formula though.

 

[evinda]

Good! ;)
No need. It suffices that you are sure that your formula is correct.
You should write down the steps how you got to the formula though.

I have already done...Thank you very much for your help! (Tongueout)

 

[alyafey22]

Use induction on $n$ to show it is correct.

 

[evinda]

Use induction on $n$ to show it is correct.

So,for $a=0$ ,we know that it is true.
We suppose that it is true for $a$.
Then I write the equation for $a+1$ ,but how can I continue??

 

[alyafey22]

Using induction on $n$ implies that we have to prove that

1- For $$n=0$$ we have

$$I_{0}=\frac{0!I_{0}}{s^{0}}=I_0$$ .

2- Assume that

$$I_{n}=e^{-sa}\sum_{k=1}^{n} \frac{a^{n+1-k}}{s^{k}}\frac{n!}{(n-k+1)!}+\frac{n!I_{0}}{s^{n}} $$

is true and prove that $$I_{n+1}=e^{-sa}\sum_{k=1}^{n+1} \frac{a^{n+2-k}}{s^{k}}\frac{(n+1)!}{(n-k+2)!}+\frac{(n+1)!I_{0}}{s^{n+1}}$$

It might be difficult but it is a really good exercise.

 

[evinda]

Using induction on $n$ implies that we have to prove that

1- For $$n=0$$ we have

$$I_{0}=\frac{0!I_{0}}{s^{0}}=I_0$$ .

2- Assume that

$$I_{n}=e^{-sa}\sum_{k=1}^{n} \frac{a^{n+1-k}}{s^{k}}\frac{n!}{(n-k+1)!}+\frac{n!I_{0}}{s^{n}} $$

is true and prove that $$I_{n+1}=e^{-sa}\sum_{k=1}^{n+1} \frac{a^{n+2-k}}{s^{k}}\frac{(n+1)!}{(n-k+2)!}+\frac{(n+1)!I_{0}}{s^{n+1}}$$

It might be difficult but it is a really good exercise.

Ok,thank you!