MHB Integral from negative infinity to infinity

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The integral from negative infinity to infinity of the function $\frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}$ evaluates to $\pi$. This result is significant in the context of complex analysis and residue theory, where such integrals often arise in evaluating contour integrals. The polynomial in the denominator can be analyzed for its roots to determine the behavior of the integrand. The discussion emphasizes the importance of understanding the properties of the function and the application of integration techniques. This integral exemplifies how advanced calculus can yield elegant results in mathematical analysis.
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$\displaystyle (1)\;\; \int_{-\infty}^{\infty}\frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}dx$
 
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\[\int_{-\infty}^{\infty}\frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}dx=\pi\]

Now can you tell us in what context this arises.

CB
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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