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Determine all $a$ and $b$ for which the this converges:

$$\int_0^{\pi/2}\frac{\mathrm{d}x}{(\sin^ax^2)(\cos^bx)}$$

I'm not sure how to approach the problem. I've tried using the comparison test, setting:

$$f(x)=1/(sin^ax^2)(cos^bx)$$ $$g(x)=1/(sin^ax^2)$$ then doing $$\lim_{x \to 0}f(x)/g(x)dx = 1/(cos^bx) = 1$$

then by comparison test I should get that whenever g(x) converges, so does f(x) -> but I'm not sure where to continue from here

Ideally I would like to show that g(x) converges whenever a = ??? and then I can also solve for b

Some help or a general direction would be appreciated note that I saw this question: For which values of $p$ and $q$ does an improper integral converge However this one is different, and I did not understand how to apply the answers given there

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$f(a, b) =\int_0^{\pi/2}\frac{\mathrm{d}x}{(\sin^ax^2)(\cos^bx)} $.

Around $x=0$, $\dfrac1{\sin^ax^2} \approx \dfrac1{x^{2a}} =x^{-2a} $ and $\dfrac1{\cos^bx} \approx 1 $, so the integral is like $\lim_{r \to 0}\int_r^c x^{-2a}dx =\lim_{r \to 0}\dfrac{x^{-2a+1}}{-2a+1}|_r^c $ and this diverges for $-2a+1 < 0$ or $a > \frac12$.

Similarly, around $x = \pi/2$, $\cos(\pi/2-x) =\sin(x) \approx x$ so so the integral is like $\lim_{r \to 0}\int_r^c x^{-b}dx =\lim_{r \to 0}\dfrac{x^{-b+1}}{-b+1}|_r^c $ and this diverges for $b > 1$.

You can show that the integral converges for $a < \frac12$ and $b < 1$.

For $a=\frac12$ and $b = 1$ the integral behaves like $\lim_{r \to 0}\int_r^c \dfrac{dt}{t} =\lim_{r \to 0}\ln(t)|_r^c $ and this diverges.

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