Integral of Rational Function with Zero Result

In summary, the integral in question can be evaluated using an indirect method by considering the function f(x) and its anti-derivatives. By choosing appropriate constants, the value of the integral can be shown to be equal to zero.
  • #1
StatusX
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I've just found that, for all a>0:

[tex] \int_0^\infty \frac{a (x^2 - 1)^2 - 2 x (x + a)^2}{(x + a)^3 (a x + 1)^3} dx = 0[/tex]

This can be found by brute force, but there must be a simpler way to show it.
 
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  • #2
StatusX said:
I've just found that, for all a>0:

[tex] \int_0^\infty \frac{a (x^2 - 1)^2 - 2 x (x + a)^2}{(x + a)^3 (a x + 1)^3} dx = 0[/tex]

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

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
Id be interested to hear any thoughts you have on it.
 
  • #4
The integrand happens to be an exact derivative of x*P(x)/Q(x) for some P and Q.
 
  • #5
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 [itex]f[/itex] be defined such that

[tex]f(x) = \frac{a(x^2-1)^2-2x(x+a)^2}{(x+a)^3(ax+1)^3}[/tex]​

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

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

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

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

[tex]h[g(x)] \leq f(x) \leq k[g(x)][/tex]​

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

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

Combining these two results, we find that ...

[tex]\int_0^{\infty}f(x)\mathrm{d}x = 0[/tex]​
 
  • #6
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
StatusX said:
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.

That's very likely. If you read the "Read This First" segment of my post, I acknowledged that.
 
  • #8
StatusX said:
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.

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
jgens said:
Since it's easy to verify that [itex]G(0) = 0[/itex] (neglecting the constant)
"Neglecting" the constant? :confused: 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)


Since [itex]lim_{x \to \infty}H(x) = 0[/itex] (once again, neglecting the constant) where [itex]H[/itex] is an anti-derivative of [itex]h[/itex], this suggests that [itex]\lim_{x \to \infty}F(t) = 0[/itex].
Again with the One True Antiderivative thing. You can certainly choose the constant so that [itex]\lim_{x \to \infty}F(t) = 0[/itex].


Of course, there's no reason those two choices have to be consistent. :frown:
 
  • #10
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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Related to Integral of Rational Function with Zero Result

1. Why is this integral zero?

The integral is zero because the function being integrated has an equal amount of positive and negative values, resulting in a cancellation effect.

2. What does it mean when an integral is equal to zero?

When an integral is equal to zero, it means that the area under the curve is equal to zero. This could indicate that the function has an equal amount of positive and negative values, or that the limits of integration result in a net zero area.

3. Is there a specific method for solving integrals that result in zero?

There is no specific method for solving integrals that result in zero. However, it is important to choose appropriate limits of integration and carefully evaluate the function being integrated to determine if it has a zero net area.

4. Can an integral be zero if the function being integrated is not symmetrical?

Yes, an integral can be zero even if the function being integrated is not symmetrical. This can occur if the function has an equal amount of positive and negative values, or if the limits of integration result in a net zero area.

5. How does the shape of the curve affect the value of the integral?

The shape of the curve can affect the value of the integral by determining the net area under the curve. For example, a function with a steep curve may have a larger net area, resulting in a non-zero integral, while a function with a flatter curve may have a smaller net area, resulting in a zero integral.

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