Proving Induction and Integral

[Benny]

Hi, I'm having trouble with the following question.

Q. Prove that

$$\int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{n - 1} } {f\left( {x_n } \right)} } } dx_n ...dx_2 dx_1 = \frac{1}{{\left( {n - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt$$

The hint says to use induction. So I tried showing that it's true for n = 1. For that particular case the statement is a single integral with terminals 0 (lower) and x (upper) I think. But that doesn't matter too much for my purposes. After assuming that the result is true for some integer k, I'm having trouble verifying the result for n = k + 1. I wrote down

$$I = \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)dx_{k + 1} dx_k } } } } ...dx_2 dx_1$$...(1)

Since the result that I've assumed has f(x_k) as the integrand rather than
f(x_(k+1)), I tried to use parts to get rid of the f(x_(k+1)). I did this by differenting f(x_(k+1)) and integrating "1". But doing this leads to integrals which cancel, leading me back to equation (1) above. Can someone help me out? Thanks.

 

[HallsofIvy]

Hi, I'm having trouble with the following question.

Q. Prove that

$$\int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{n - 1} } {f\left( {x_n } \right)} } } dx_n ...dx_2 dx_1 = \frac{1}{{\left( {n - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt$$

The hint says to use induction. So I tried showing that it's true for n = 1. For that particular case the statement is a single integral with terminals 0 (lower) and x (upper) I think. But that doesn't matter too much for my purposes.

Yes, in case n= 1, that becomes
[tex]\int_0^x f(x)dx= \int_0^x f(t)dt[/itex]
which is true since it is the same integral with different "dummy" variables.

After assuming that the result is true for some integer k, I'm having trouble verifying the result for n = k + 1. I wrote down

$$I = \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)dx_{k + 1} dx_k } } } } ...dx_2 dx_1$$...(1)

Since the result that I've assumed has f(x_k) as the integrand rather than
f(x_(k+1)), I tried to use parts to get rid of the f(x_(k+1)). I did this by differenting f(x_(k+1)) and integrating "1". But doing this leads to integrals which cancel, leading me back to equation (1) above. Can someone help me out? Thanks.

So your k+1 integral is just the k integral with
$$fx_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}$$.
How about integration by parts with
$$u=f_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}$$.
and dv= dx_k?

 

[Benny]

I'm not sure what you mean by

$$fx\left( {x_k } \right) = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1}$$

Is f applied to x_k or x(x_k)?

Your suggestion looks like what I tried before.

$$\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} = \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \int\limits_0^{x_k } {x_{k + 1} f'\left( {x_{k + 1} } \right)dx_{k + 1} }$$

$$= \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } + \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1}$$

Which is the "n = k+1" integral I started with. Unless I missed something.

 

[HallsofIvy]

Sorry, I had an extra "x" in that formula!

No, my suggestion is not exactly what you did before. You tryed to integrate
$$\int_0^{x_k}f(x_{k+1})dx_{k+1}$$
by parts. My suggestion was to integrate
$$\int_0^{x_{k-1}}\int_0^{x_k}f(x_{k+1})dx_{k+1}dx_k$$
taking
$$u= \int_0^{x_k}f(x_{k+1})$$
and du= dx_k+1.

(If you use integration by parts twice swapping what you use for u and dv in the middle, of course you just get back to what you started with- I'm saying use integration by parts once to get rid of that innermost integral.)

Look at a simple example:
$$\int_0^x\int_0^y f(t)dt dy$$
Let $u= \int_0^y f(t)dt$ and dv= dy. Then du= f(y) and v= y. We have
$$uv\|_0^x- \int_0^xvdu= y\int_0^xf(t)dt- \int_0^x yf(y)dy$$
If we change the dummy variable y in the second integral to t, we have
$$y\int_0^xf(t)dt- \int_0^x tf(t)dt= \int_0^x(y-t)f(t)dt$$
Do you see how that fits into your induction?

 

[Benny]

Oh ok, I can see that what I did was different to what you suggested. I misread it before.

I'm not aware of any other requirements. The question doesn't refer to other conditions which are required.

I'll see what I can come up with, thanks.

 

[Benny]

I've thought about this and I still can't get it out.

Firstly, if $u = \int\limits_0^y {f\left( t \right)dt} ,dv = dy$ then $du = f\left( y \right)dy,v = y$ by FTC. Ok I get that part. But what about uv evaluated at 0 and x?

In the example you gave, the outer most integral was done wrty so I would've thought that would mean in evaluating uv at 0 and x, where the variable 'y' appear in uv, it would be replaced by x. That is,

$$\left[ {uv} \right]_0^x = \left[ {y\int\limits_0^y {f\left( t \right)dt} } \right]_0^x = x\int\limits_0^x {f\left( t \right)dt}$$

In any case, considering the following integral "I" and integrating by parts as you suggested, with

$$u = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} ,dv = dx_k$$ I obtain:

$$I = \int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } dx_{k + 1} dx_k$$

$$= \left[ {x_k \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } \right]_0^{x_{k - 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)} dx_k$$

$$= x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_{k + 1} } \right)dx_{k + 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k }$$

$$= x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_k } \right)dx_k } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k }$$

$$= \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)f\left( {x_k } \right)dx_k }$$

I'm not sure how I can use the above and the assumption (that the statement is true for n = k) to show that the n = k + 1 case is true.

 

[HallsofIvy]

Remember that you can change "dummy" variables at will. You need to change that x_k+1 to "t".

 

[Benny]

I can rewrite the last line as:

$$\int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)} f\left( {x_k } \right)dx_k = \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - t} \right)} f\left( t \right)dt$$

But I don't see how I can use the the assumption to prove the result from the above.

 

[HallsofIvy]

All those x_k are confusing the devil out of me! That's why earlier I gave an example with x and y. I showed earlier that
$$\int_0^x\int_0^y f(t)dt dy= \int_0^x(y-t)f(t)dt$$
That can be put in terms of x_k and x_k+1 by simply make x= x_k-1 and y= x_k:
$$\int_0^{x_{k-1}}\int_0^{x_k}f(t)dt dy= \int_0^{x_{k-1}(x_k-t)f(t)dt$$

Now apply your induction hypothesis, that
$$\int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {f\left( {x_n } \right)} } } dx_k ...dx_2 dx_1 = \frac{1}{{\left( {k - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt$$
with f(x) replaced by (x_k-x)f(x).

 

[Benny]

Ok I'll try that thanks.