# Why is this integral zero

1. Aug 18, 2010

### StatusX

I've just found that, for all a>0:

$$\int_0^\infty \frac{a (x^2 - 1)^2 - 2 x (x + a)^2}{(x + a)^3 (a x + 1)^3} dx = 0$$

This can be found by brute force, but there must be a simpler way to show it.

2. Aug 18, 2010

### jgens

How formal does the solution need to be? I have an idea that might work out, but I don't know if it would be exactly what you're looking for.

3. Aug 18, 2010

### StatusX

Id be interested to hear any thoughts you have on it.

4. Aug 18, 2010

### hamster143

The integrand happens to be an exact derivative of x*P(x)/Q(x) for some P and Q.

5. Aug 18, 2010

### jgens

Read This First: Here are my first thoughts. The biggest problem with this is that it's not rigorous and utilizes a lot of 'hand-waving arguments' (and I'm not positive that this hand-waving is even justified here!). Another problem with it is that it's not really all that simple, so even if it can be justified, I don't think that it's much good as a solution. I think that hamster143 outlined a much better solution too. But with that qualified, here was my first thought about the problem.

Let the function $f$ be defined such that

$$f(x) = \frac{a(x^2-1)^2-2x(x+a)^2}{(x+a)^3(ax+1)^3}$$​

Since $f$ is continuous on $[0,\infty)$, we can apply the second fundamental theorem of calculus to find that ...

$$\int_0^{\infty}f(x)\mathrm{d}x = \lim_{x \to \infty}F(t) - F(0)$$​

where $F$ is an anti-derivative of $f$. Therefore, we need only find the values of these anti-derivatives. We'll go about this in an indirect way.

To find $F(0)$, first define the function $g$ such that $g(x) = a(x^2-1)^2-2x(x+a)^2$. Next, note that by choosing $x$ small enough, we can find numbers $h,k > 0$ such that the following inequality holds:

$$h[g(x)] \leq f(x) \leq k[g(x)]$$​

Since it's easy to verify that $G(0) = 0$ (neglecting the constant) where $G$ is an anti-derivative of $g$, this suggests that the anti-derivative of $f$ at zero is equal to zero. Therefore, $F(0) = 0$.

To evaluate the term $\lim_{x \to \infty}F(t)$, we can note that as $x$ becomes arbitrarily large, $f(x)$ tends to something like $h(x) = x^{-2}$ (because the numerator is a polynomial of degree 4 and the denominator is a polynomial of degree 6). Since $lim_{x \to \infty}H(x) = 0$ (once again, neglecting the constant) where $H$ is an anti-derivative of $h$, this suggests that $\lim_{x \to \infty}F(t) = 0$.

Combining these two results, we find that ...

$$\int_0^{\infty}f(x)\mathrm{d}x = 0$$​

6. Aug 19, 2010

### StatusX

hamster,

I'm not sure what you mean. Are P(x) and Q(x) supposed to be polynomials? According to mathematica, the indefininte integral contains logs, and is very complicated.

jgens,

I'm not sure I understand exactly what you're saying, but I'm skeptical of your method since you don't seem to be using the specific form of the function I gave. For example, if I changed the (ax+1) in the denominator to (ax+2), the integral would no longer be zero, but I'm not sure your argument would be any different. I think you're neglecting constants when you shouldn't be.

7. Aug 19, 2010

### jgens

That's very likely. If you read the "Read This First" segment of my post, I acknowledged that.

8. Aug 19, 2010

### hamster143

Not according to wolfram alpha ...

http://www.wolframalpha.com/input/?i=\int+\frac{a+%28x^2+-+1%29^2+-+2+x+%28x+%2B+a%29^2}{%28x+%2B+a%29^3+%28a+x+%2B+1%29^3}+dx+

9. Aug 19, 2010

### Hurkyl

Staff Emeritus
"Neglecting" the constant? You make it sound like there's One True Antiderivative which all others are just modifications of!

Of course, you can choose the constant so that G(0) = 0. And you can choose the corresponding constant so that F(0) = 0.
(it would be easier to choose F(0) directly)

Again with the One True Antiderivative thing. You can certainly choose the constant so that $\lim_{x \to \infty}F(t) = 0$.

Of course, there's no reason those two choices have to be consistent.

10. Aug 19, 2010

### JJacquelin

A method which doesn't require heavy developments is given in attachment.

( Typo in the attachment : remplace equation -Aa=-a by -Aa=a )

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Last edited: Aug 19, 2010