Integral Proof (1 Viewer)

shongaponga · Aug 23, 2012

Hey can someone please help me with this proof that I'm having trouble with, any help is appreciated

nightweaver066 · Aug 23, 2012

Numerator becomes $x(x^2 - 1) + x^2 - 1$

Split the fraction in to two, factorise denominator as difference of two squares and simplify.

One function is odd so that becomes 0, other one becomes $tan^{-1}x$

As x -> infinity, $tan^{-1}x \rightarrow \frac{\pi}{2}$

As x-> -infinity, $tan^{-1}x \rightarrow -\frac{\pi}{2}$

Evaluating the integral, you get $\pi$

shongaponga · Aug 23, 2012

Oh god how didn't I see that lol, thanks!

Trebla · Aug 23, 2012

Being somewhat 'rigorous', first split the integral first to account for the undefined bits

$\displaystyle\int_{-\infty}^{-1}\dfrac{x^3+x^2-x-1}{x^4-1}\,dx + \displaystyle\int_{-1}^1\dfrac{x^3+x^2-x-1}{x^4-1}\,dx + \displaystyle\int_1^{\infty}\dfrac{x^3+x^2-x-1}{x^4-1}\,dx$

First note that

$\displaystyle\int\dfrac{x^3+x^2-x-1}{x^4-1}\,dx\\\\=\displaystyle\int\dfrac{x(x^2-1)+(x^2-1)}{(x^2-1)(x^2+1)}\,dx\\\\=\displaystyle\int\dfrac{x+1}{x^2+1}\,dx\\\\=\dfrac{1}{2}\ln(x^2+1) + \tan^{-1}x + c$

So applying this result to the definite integrals we get

$\displaystyle\lim_{a\rightarrow - 1}\left(\dfrac{1}{2}\ln(a^2+1) + \tan^{-1}a\right) - \displaystyle\lim_{a\rightarrow-\infty} \left(\dfrac{1}{2}\ln(a^2+1)+\tan^{-1}a\right) + \displaystyle\lim_{a\rightarrow 1}\left(\dfrac{1}{2}\ln(a^2+1) + \tan^{-1}a\right) - \displaystyle\lim_{a\rightarrow-1} \left(\dfrac{1}{2}\ln(a^2+1)+\tan^{-1}a\right)+\displaystyle\lim_{a\rightarrow \infty}\left(\dfrac{1}{2}\ln(a^2+1) + \tan^{-1}a\right) - \displaystyle\lim_{a\rightarrow 1} \left(\dfrac{1}{2}\ln(a^2+1)+\tan^{-1}a\right)\\\\=\displaystyle\lim_{a\rightarrow \infty}\left(\dfrac{1}{2}\ln(a^2+1) + \tan^{-1}a\right) - \displaystyle\lim_{a\rightarrow-\infty} \left(\dfrac{1}{2}\ln(a^2+1)+\tan^{-1}a\right)$

But (not sure if this is well justifed)

$\displaystyle\lim_{a\rightarrow-\infty}\ln(a^2+1)=\displaystyle\lim_{a\rightarrow \infty}\ln(a^2+1)$

Hence we are left with

$\displaystyle\lim_{a\rightarrow \infty}\tan^{-1}a - \displaystyle\lim_{a\rightarrow-\infty}\tan^{-1}a$

which equals

$\pi$

The reason I wouldn't evaluate that integral directly is because the function is undefined at x = -1 and x = 1 so you need to separate the different cases and test for convergence. There are examples where if you integrate across a point which is not defined for the integrand, you get an incorrect answer.

shongaponga · Aug 23, 2012

Thanks Trebla

Carrotsticks · Aug 23, 2012

http://en.wikipedia.org/wiki/Cauchy_principal_value

math man · Aug 24, 2012

ahh my trial question, nice

seanieg89 · Aug 25, 2012

The 1 and -1 are no problem here as they are removable singularities. It is understood that you just give the integrand the value that would make it continuous there.

In fact you could take a continuous function on the interval [a,b] and change it at a countable number of points without changing the fact that it is Riemann integrable and without changing the value of its integral.

That said, I do not like the wording of the question because it is essentially asking students to find the Cauchy Principal Value of the given integral without explicitly saying so. As written, the improper integral is not well defined. Eg if we integrated from -a to 2a and let a->inf we would get a different value for our integral, which would not happen with a convergent improper integral on R.

Integral Proof (1 Viewer)

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