# Trouble with an integral

1. Mar 31, 2007

### gnome

This almost seems to easy to be true. Is there anything wrong with this:

$\mbox{Find} \int_0^1 x^p\, (ln\, {x})^3 \, dx$

$\mbox{First let } F(p) = \int_0^1 x^p\,dx = \frac{x^{p+1}}{p+1} |_0^1 = \frac{1}{p+1}$

$\frac{\partial}{\partial p}( \frac{1}{p+1}) = \frac{-1}{(p+1)^2}$

$\mbox{then }F'(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}$

$\mbox{so now let } G(p) = \int_0^1 x^p\,ln\,x\,dx = \frac{-1}{(p+1)^2}$

$\frac{\partial}{\partial p} (\frac{-1}{(p+1)^2}) = \frac{2}{(p+1)^3}$

$\mbox{then }G'(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3}$

$\mbox{Finally let } H(p) = \int_0^1 x^p\,(ln\,x)^2\,dx = \frac{2}{(p+1)^3}$

$\frac{\partial}{\partial p}( \frac{2}{(p+1)^3}) = \frac{-6}{(p+1)^4}$

$\mbox{therefore } H'(p) = \int_0^1 x^p\,(ln\,x)^3\,dx = \frac{-6}{(p+1)^4}$

Last edited: Mar 31, 2007
2. Mar 31, 2007

### quasar987

I don't understand line 4.

3. Mar 31, 2007

### VietDao29

p in the problem is a constant, and dx is your variable of integration, and therefore you should not define $$F(\fbox{p}) = \mathop \int \limits_0 ^ 1 x ^ p dx$$
A normal Integration by Parts will work. Let:
u = ln3x => du = 3 ((ln2x) / x) dx
dv = xpdx => v = xp + 1 / (p + 1)
$$\mathop \int \limits_0 ^ 1 x ^ p \ln ^ 3 x dx = \left. \frac{x ^ {p + 1} \ln ^ 3 x}{p + 1} \right|_0 ^ 1 - \frac{3}{p + 1} \int \limits_0 ^ 1 x ^ p \ln ^ 2 x dx$$.
Can you go from here? :)

4. Mar 31, 2007

### gnome

If A = B, dA/dp = dB/dp.

Here, $\mbox{A }= F(p) = \int_0^1 x^p\,dx$
and
$\mbox{B } = \frac{1}{p+1}$

And the way I defined F(p), the value of F'(p) is the solution to the original problem.

Why not? Why should I not define any "helper" function I want as long as I don't do anything illegal to the original function? p is not necessarily a constant; it's just not the variable of integration. Think of it as $E(p,x) = x^p\,ln^3\,x$, and the original question as $\int_{a(p)}^{b(p)}\,E(p,x)\,dx$

Yes, it can be solved using integration by parts 3 times, doing 3 logarithmic integrals (and giving the same answer), but isn't this Leibniz Rule method much more elegant?

Last edited: Mar 31, 2007
5. Mar 31, 2007

### quasar987

I should have been more specific. It is

$$F'(p)=\int_0^1x^p\ln(x)dx$$

that I don't get. To me it would be

$$F'(p)=\int_0^1px^{p-1}dx$$

6. Mar 31, 2007

### gnome

You're taking
$F(x) = x^p\,dx$ where p is a constant and differentiating wrt x:
$F'(x) = px^{p-1}$ but you can't do that through the integral sign.
Remember that this is a definite integral so $\int_0^1 x^p\,dx$ is a NUMBER.
If you say $F(x) = \int_0^1 x^p\,dx$, it's like saying $F(x) = 5$ (just picking a number out of the air), but with confusing notation since your x is being used two different ways.
Then $F'(x) = \frac{dF}{dx} = 0$.

I'm making the integrand a function of p and x [call it E(p,x)] and defining the entire rhs (the integral of E) as a function of p, so when I differentiate through the integral sign all I get for F'(p) is the partial derivative of E wrt p (since the limits of integration are both constants), which is equivalent to
$F(p) = x^p$
$F'(p) = x^p\,ln(x)$ (remember, logarithmic differentiation.)

The general rule is here:
http://mathworld.wolfram.com/LeibnizIntegralRule.html

Last edited: Mar 31, 2007
7. Mar 31, 2007

### arildno

It is certainly admissible to form an auxiliary function with its independent variable being p.

The only thing that might have invalidated the approach would have been where you switch the order of the limiting operations "integration" and "differentiation".

"Normally", however, as in your case, this switch is allowed.

8. Mar 31, 2007

### arildno

For those still skeptical of gnome's result, let us consider the sequence of integrals:
$$I_{n}=\int_{0}^{1}x^{p}(\ln(x))^{n}dx}$$
Integration by parts yields, for p>-1:
$$I_{n}=\frac{x^{p+1}}{p+1}(\ln(x))^{n}\mid_{x=0}^{x=1}-\frac{n}{p+1}\int_{0}^{1}x^{p}(\ln(x))^{n-1}dx=-\frac{n}{p+1}I_{n-1}$$

That is, we have the recursion relation:
$$I_{n}=-\frac{n}{p+1}I_{n-1}$$
which is, precisely, gnome's result, remembering that:
$$I_{0}=\frac{1}{p+1}$$

9. Mar 31, 2007

### gnome

Please elaborate on what you mean by that. I don't want to inadvertently get into trouble.

10. Mar 31, 2007

### arildno

Let us have an integral of the form:
$$I(\alpha)=\int_{a}^{b}f(x,\alpha)dx$$
The derivative of I with respect to alpha is, as usual, given by the expression:
$$I'(\alpha)=\lim_{h\to{0}}\frac{I(\alpha+h)-I(\alpha)}{h}=\lim_{h\to{0}}\int_{a}^{b}\frac{f(x,\alpha+h)-f(x,\alpha)}{h}dx$$
where I've used additivity of integrals, among other manipulations.

Now, this limit may well exist, even if, say, the partial derivative of f with respect to alpha does NOT exist.

That is, there exists cases where the above expression is valid, whereas the switched version,
$$\int_{a}^{b}(\lim_{h\to{0}}\frac{f(x,\alpha+h)-f(x,\alpha)}{h})dx$$
does not exist, for example if the integrand is undefined.

11. Mar 31, 2007

### gnome

If I understand you correctly, in those instances where the partial derivative of f with respect to alpha doesn't exist, this method of finding the definite integral is invalid, so in order to use it one must be certain that the partial derivative is defined. Correct?

But are there instances where the partial derivative IS defined but the integral still cannot be obtained by this method?

Oh, and does the derivative have to be defined everywhere, or is it sufficient that it be defined over the range of the integral?