What is a good way to see that the integral $$ \int_{0}^{\pi} | \cos x |^{-t} dx $$ exists for $t \in (-\infty,1)$ and diverges for $t \in [1,\infty)$?
When does the integral $\int_{0}^{\pi} | \cos x |^{-t} dx$ exist?
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calculus
real-analysis
integration
analysis
definite-integrals
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0The midpoint of the integration range might be a singular point. You just have to discuss in which cases it really is, and it leads to an integrable singularity or don't. – 2017-01-28
2 Answers
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HINT:
$\cos(x)=(\pi/2-x)+O(\pi/2-x)^2$. For what values of $t$ does $\int_0^\pi \frac{1}{(\pi/2-x)^t}\,dx$ converge?
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$$\int_0^\pi|\cos x|^{-t}dx=2\int_0^{\frac{\pi}{2}}\sin^{-t}xdx\leq\int_0^{\frac{\pi}{2}}x^{-t}dx=-\frac{2}{t+1}\Big(x^{1-t}\Big)_0^{\frac{\pi}{2}}$$ is finite where $t<1$. $$\int_0^\pi|\cos x|^{-t}dx=2\int_0^{\frac{\pi}{2}}\sin^{-t}xdx\geq\int_0^{\frac{\pi}{2}}(\dfrac{x}{2})^{-t}dx=-\frac{2^t}{t+1}\Big(x^{1-t}\Big)_0^{\frac{\pi}{2}}=\infty$$ where $t>1$. For $t=1$ the second shows $$\int_0^\pi|\cos x|^{-t}dx\geq\int_0^{\frac{\pi}{2}}(\dfrac{x}{2})^{-t}dx=2^t\Big(\ln x\Big)_0^{\frac{\pi}{2}}=\infty$$
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0For $x\ge 0$, $\sin(x)\le x\implies \frac{1}{\sin(x)}\ge \frac1x$. So, the first argument is flawed. It can be corrected by noting that for $0\le x\le \pi/2$, $\sin(x) \ge \frac{2x}{\pi}$. – 2017-01-28