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

[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 \(\displaystyle n=0\) we have

\(\displaystyle I_{0}=\frac{0!I_{0}}{s^{0}}=I_0\) .

2- Assume that

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

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

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 \(\displaystyle n=0\) we have

\(\displaystyle I_{0}=\frac{0!I_{0}}{s^{0}}=I_0\) .

2- Assume that

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

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

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

Ok,thank you!