It is a exercise from Calculus appeared in improper integral, one of my classmates ask me. It reads as follows:
$\int_{0}^{\infty} \log(1-\frac{a}{x^2}+\frac{1}{x^4}) dx = ?$, where a is a parameter and a<2.
I have tried to let t = $\frac{1}{x}$ and rewrite LHS as $\int_{0}^{\infty} \frac{\log(1-\frac{a}{x^2}+\frac{1}{x^4})}{x^2} dx$. Or write it as $\int_{0}^{\infty} \log(x^4 - ax^2 +1) dx$ $-\int_{0}^{\infty} 4\log(x) dx$ in a more simple form. But I still cannot calculate it directly. I wonder whether we should use more delicate techniques to solve the problem?
I know it does not contain any philosophy in this problem but I would like to regard it as an intelligent challenge.
I would appreciate it if anyone can give me some insights.
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Edward Z. Miao
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What is $a$. For large values of $a$ the integrand is not well defined. – Kavi Rama Murthy Mar 14 '19 at 06:45
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$\int_0^\infty log(x)dx$ doesn't exist. – FDP Mar 14 '19 at 08:43
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If $|a|\le 2$, the parameterisation $a=2\cos 2\theta$ obtains the answer as $2\pi\sin\theta$, i.e. $\pi\sqrt{2-a}$. If $a>2$, there are real values of $a$ for which the logarithm's argument is negative, so that the integrand is non-real. The former case can presumably be analytically continued to $a<-2$.
J.G.
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See my answer here: https://math.stackexchange.com/a/3144140/186817
(replace $a$ by $-\frac{a}{2}$ into the formula)
The desired result is $\pi\sqrt{2-a}$
(Observe that one needs $a\leq 2$ )
FDP
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