»Prove that \(\displaystyle \int_{1}^{\infty}\frac{dx}{x\sqrt{|2-x^2|}}\) a) exists, b) is convergent and c) solve it.«
a): \(\displaystyle \frac{1}{x\sqrt{|2-x^2|}}\) is less* than \(\displaystyle \frac{1}{x^2}\), whose integral \(\displaystyle \int_{1}^{\infty}\frac{dx}{x^2}\) exists. Therefore, \(\displaystyle \int_{1}^{\infty}\frac{dx}{x\sqrt{|2-x^2|}} < \int_{1}^{\infty}\frac{dx}{x^2} < \infty\).
b): what kind of questions is this? If it exists, it is convergent. Or, if it is not convergent, it does not exist! (OK, what am I missing here?
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c): \(\displaystyle =\int_{1}^{\sqrt2}\frac{dx}{x\sqrt{2-x^2}}+\int_{\sqrt2}^{\infty}\frac{dx}{x\sqrt{2-x^2}}=\lim_{\epsilon\to\sqrt2}\int_{1}^{\epsilon}\frac{dx}{x\sqrt{2-x^2}}+\lim_{\epsilon\to\sqrt2}\int_{\epsilon}^{\infty}\frac{dx}{x\sqrt{-(2-x^2)}}=\) :?:
*Is this claim valid? It seems "obvious", but what about proving it (ie. showing that the difference between the later and the former is positive)? I'd go: (for \(\displaystyle x<\sqrt2\)) \(\displaystyle \frac{1}{x^2}-\frac{1}{x\sqrt{2-x^2}}=\frac{\sqrt{2-x^2}-x}{x^2\sqrt{2-x^2}}=\) :?: Again, it seems "obvious": denominator is always positive, and so is the nominator as we have something slightly greater than x (because of the \(\displaystyle 2\) added under the \(\displaystyle \sqrt\)). What about (for \(\displaystyle x>\sqrt2\)) :?: Here we'd get \(\displaystyle \frac{1}{x^2}-\frac{1}{x\sqrt{-(2-x^2)}}=\frac{1}{x^2}-\frac{1}{xi\sqrt{(2-x^2)}}\) ... what to do with imaginary number?
a): \(\displaystyle \frac{1}{x\sqrt{|2-x^2|}}\) is less* than \(\displaystyle \frac{1}{x^2}\), whose integral \(\displaystyle \int_{1}^{\infty}\frac{dx}{x^2}\) exists. Therefore, \(\displaystyle \int_{1}^{\infty}\frac{dx}{x\sqrt{|2-x^2|}} < \int_{1}^{\infty}\frac{dx}{x^2} < \infty\).
b): what kind of questions is this? If it exists, it is convergent. Or, if it is not convergent, it does not exist! (OK, what am I missing here?
)c): \(\displaystyle =\int_{1}^{\sqrt2}\frac{dx}{x\sqrt{2-x^2}}+\int_{\sqrt2}^{\infty}\frac{dx}{x\sqrt{2-x^2}}=\lim_{\epsilon\to\sqrt2}\int_{1}^{\epsilon}\frac{dx}{x\sqrt{2-x^2}}+\lim_{\epsilon\to\sqrt2}\int_{\epsilon}^{\infty}\frac{dx}{x\sqrt{-(2-x^2)}}=\) :?:
*Is this claim valid? It seems "obvious", but what about proving it (ie. showing that the difference between the later and the former is positive)? I'd go: (for \(\displaystyle x<\sqrt2\)) \(\displaystyle \frac{1}{x^2}-\frac{1}{x\sqrt{2-x^2}}=\frac{\sqrt{2-x^2}-x}{x^2\sqrt{2-x^2}}=\) :?: Again, it seems "obvious": denominator is always positive, and so is the nominator as we have something slightly greater than x (because of the \(\displaystyle 2\) added under the \(\displaystyle \sqrt\)). What about (for \(\displaystyle x>\sqrt2\)) :?: Here we'd get \(\displaystyle \frac{1}{x^2}-\frac{1}{x\sqrt{-(2-x^2)}}=\frac{1}{x^2}-\frac{1}{xi\sqrt{(2-x^2)}}\) ... what to do with imaginary number?