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I wish to check if the integral:

$$\int_0^{\frac{\pi}{2}}\sin (\frac{1}{\cos (x)})$$

is converging.

Here is what I did:

$\int_0^{\frac{\pi}{2}}\sin (\frac{1}{\cos (x)})\lt \int_0^{\frac{\pi}{2}}\sin (\frac{1}{1})=\sin (1)\frac{\pi}{2}$ so the integral converge.

Is this true?

segevp
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2 Answers2

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The idea might be correct, but what you wrote certainly is not, since $$\sin\left(\frac{1}{\cos x}\right)<\sin (1)$$ is not true.

It is not true for example for $$x=\arccos\frac{2}{\pi}$$ in which case $$\sin\left(\frac{1}{\cos x}\right)=\sin(\frac\pi2)=1>\sin(1)$$

5xum
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As

$$-1 \le \sin(x) \le +1$$

we get

$$-\int_0^{\pi/2} dx \le \int_0^{\pi/2} \sin\left(\frac{1}{\cos(x)}\right) dx \le \int_0^{\pi/2} dx$$

so

$$-\frac{\pi}2 \le \int_0^{\pi/2} \sin\left(\frac{1}{\cos(x)}\right) dx \le \frac{\pi}2$$

johannesvalks
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    thank you. nice answer – segevp Jul 18 '17 at 11:44
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    Not nice answer, no. $\int_0^X \sin(t) dt = 1-\cos(X)$, so $0\le \int_0^X \sin(t) dt \le 2$, but that doesn't mean $\int_0^{+\infty} \sin(t)dt$ is convergent ! For example, what sense do you give to $\int_0^1 \chi(t)dt$, where $\chi(t)=1$ if $t\in\mathbb Q$, else $0$ ? – Nicolas FRANCOIS Jul 18 '17 at 12:54
  • @NicolasFRANCOIS, $-X \le \int_0^X \sin(f(x)) dx \le X$, for continues $f: \mathbb{R} \mapsto \mathbb{R}$ – johannesvalks Jul 18 '17 at 14:30
  • @NicolasFRANCOIS, given your definition of $\chi(t)$, we get $0 \le \int_0^1 \chi(t) dt \le 1$... – johannesvalks Jul 18 '17 at 14:35
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    @johannesvalks : does this mean $\chi$ is integrable ??? mathematicians had to find a VERY GOOD integral to give a sense to $\int_0^1\chi$ ! According to Riemann, this integrale DOESN'T exist ! – Nicolas FRANCOIS Jul 18 '17 at 20:11
  • @NicolasFRANCOIS, Riemann cannot - but Lebesque can ;) – johannesvalks Jul 19 '17 at 09:23
  • @NicolasFRANCOIS, look here: https://math.stackexchange.com/questions/755576/the-lebesgue-integral-and-the-dirichlet-function – johannesvalks Jul 19 '17 at 10:01
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    @johannesvalks : I know all this. I just wanted to mention that saying the function is bounded is not enough to declare it integrable. Hence the "Not nice answer" comment. Now if you accept Lebesgue as your primary tool, everything becomes much cooler. But there is a price to pay ! – Nicolas FRANCOIS Jul 19 '17 at 10:06