# Proof Involving Integration by Parts and a Series of Functions

1. May 12, 2014

### Tollschnee

1. The problem statement, all variables and given/known data
Let f be continuous on an interval I containing 0, and define f1(x) = ∫f(t)dt, f2(x) = ∫f1(t)dt, and in general, fn(x) = ∫fn-1(t)dt for n≥2. Show that fn+1(x) = ∫[(x-t)n/n!]f(t)dt for every n≥0.

ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( )

2. Relevant equations
Integration by parts: ∫u dv = uv - ∫vdu.

3. The attempt at a solution
I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into fn+1(x) = ∫[(x-t)n/n!]f(t)dt and get f1(x) = ∫f(t)dt.

Now the induction step is where I get messed up. I assume that fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(t)dt.

I have tried doing an integral by parts setting u = fn(t) ; du = fn-1(t) ; dv = dt ; and v = t... but I keep ending up with either sequences that don't mean anything or some annoying cyclical stuff where I end up right back where I started.

2. May 13, 2014

### NihilTico

I think you are on the right track.
So $f_1(x)=\int_{0}^{x}{f_0(x)}$ and $f_n(x)=\int_{0}^{x}{f_{n-1}(x)}$ and $f_{n+1}(x)=\int_{0}^{x}{f_{n}(x)}$.

Here's the key insight:
If $x$ is allowed to be any real number, then, first considering the definite integral and letting the indefinite integral of $f_{n}(t)$ be $\int{f_{n}(t)dt}=F_{n}(t)+C$, we see that $f_{n+1}(x)=\int_{0}^{x}{f_{n}(t)}=F_{n}(x)-F_{n}(0)$. This certainly looks as if it were saying that, for any value of $x$, we have that $f_{n+1}(t)=\int{f_{n}(t)dt}+F_{n}(0)$ and, hence, $f_{n+1}'(t)=f_{n}(t)$.

Once you have that, you can integrate by parts, and some things should become clear. I'll try to point you in the right direction.
Letting $f_{n+1}(x)=\int_{0}^{x}$ I integrate by parts and obtain $f_{n+1}(x)=xf_{n}(x)-\int_{0}^{x}{tf_{n}'(t)dt}$. But, don't we know another way of expressing $f_{n}(x)$? We do, in fact, and as we are given that this can be expressed as an integral of another function. What about $f_{n}'(t)$? Well, because of the insight that $f_{n+1}'(t)=f_{n}(t)$ I claim that this can be expressed as $f_{n-1}'(t)$ and that, combining these two things I have stated here, you have something that may look familiar and you should be able to follow it through from there!

3. May 13, 2014

### vela

Staff Emeritus
It's not a fact that
$$f_{n+1}(x) = \int \frac{(x-t)^n}{n!} f(t)\,dt.$$ That's what you're supposed to prove.

Another problem I see is in your definition $f_{n+1}(x) = \int f_n(t)\,dt$. The lefthand side is a function of x, but x doesn't appear anywhere on the righthand side.

4. May 13, 2014

### SammyS

Staff Emeritus
vela must have missed those limits of integration.

You could try sub/super-scripts.

0x along with larger size to make "∫" into "" .

That looks OK for the base step.
In line with what vela said: You are to prove
$\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ ,\$ not assume it.​

For the inductive step, I like to state something like:
Assume it's true that: $\displaystyle f_{k}(x) = \int_0^x \frac{(x-t)^{k-1}}{(k-1)!} f(t)\,dt\ ,\$ for some k ≥ 0. ​

From that and using algebra & calculus you need to show that you can arrive at:
$\displaystyle f_{k+1}(x) = \int_0^x \frac{(x-t)^{k}}{(k)!} f(t)\,dt\ ,\$​

This is where you are likely to use integration by parts.

Do you see that logic behind the inductive process?

In the base step, you showed the general statement is true for k = 0 ( or n=0, depending on the index you use).

If you prove the inductive step, then the statement being true for k = 0, means that you showed it's true for k = 1.

The statement being true for k = 1, means that you showed it's true for k = 2.

The statement being true for k = 2, means that you showed it's true for k = 3.

etc.

So $\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ \$ is true for all n ≥ 0 .

Last edited: May 13, 2014
5. May 14, 2014

### vela

Staff Emeritus
Sure did.