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I stumbled upon exercise $25.a$ in section $5.11$ of Tom Apostol's Calculus $($vol $1)$.

Let:

$$ f(n) = \int _0^{\pi/4}{\rm tan}^n(x) \;{\rm dx}, \quad n \ge 1 $$

The task is to prove $f(n+1) < f(n)$. It would be easy to prove, if comparison theorem 1.20 (page 81 in that book) would have $<$ sign instead of $\le$. However, he mentions explicitly that it is harder to prove strict inequality, even though it holds (i.e. if integrands are strictly smaller, then their corresponding integrals over that interval are also strictly smaller).

My question is if anyone here has a solution to that exercise without using theorem $1.20$ as if it is with strict inequality? Alternatively, is there a readable proof of that theorem with strict inequality (maybe he would expect the reader to be able to prove it by Chapter $5$)?

Thanks!

Alessio K
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S11n
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    Strict inequality can be demonstrated, via common sense, with only a little more grunt work. I am unfamiliar with the exact theorem, but I surmise that it asserts that if $f(x) \leq g(x)$ throughout $[a,b]$ then $\int_a^b f(x)dx \leq \int_a^b g(x)dx.$ Apostol's use of an infinite number of very narrow rectangles to illustrate Reimann integration should support this assertion. ...see next comment. – user2661923 Sep 17 '21 at 10:36
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    Would it be just simply because for $x \in (0,\pi/4)$ we have $0 < \tan x < 1,$ hence $$\tan^n x > \tan^{n+1} x.$$ I dunno if I'm making mistake here. – Azlif Sep 17 '21 at 10:36
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    @Azlif Yes, that is most of the point that I was going to make. However, re my last comment, I was going to say more. Given any two functions $f_n$ and $f_{n+1}$ you should be able to do two things: [1] Identify a precise subset of $[0,1]$ of a discrete width (e.g. $\pi/10$), such that $f_{n+1}(x) < f_n(x)$, throughout this discrete width. [2] Verify that there exists a discrete minimum $\left(\text{e.g.} ~ \displaystyle \frac{1}{10^n}\right)$ such that $f_n(x) - f_{n+1}(x) \geq \frac{1}{10^n}$ throughout this interval. In my opinion, this added premise gives you strict inequality. – user2661923 Sep 17 '21 at 10:42
  • Can we just use the fact that if $f$ is continuous on an interval $(a,b)$ with $f(x) > 0$ then $$\int_a^b f(x) \ dx > 0.$$ – Azlif Sep 17 '21 at 10:52
  • I knew Apostol said that the general case is harder (in page 81), but a continuity assumption is just enough. – Azlif Sep 17 '21 at 10:59
  • Azlif, how do you know that is the fact (I haven't seen that proven before)? I know it is for $\ge$, but not for $>$. – S11n Sep 18 '21 at 14:35

1 Answers1

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We have

$$f(n)-f(n+1)=\int_{0}^{\frac{\pi}{4}}\left(\tan^n(x)-\tan^{n+1}(x)\right)dx$$ $$=\int_{0}^{\frac{\pi}{4}}\tan^{n}(x)\left(1-\tan(x)\right)dx>0$$

since $0<\tan(x)<1$ for $0<x<\frac{\pi}{4},$ so the integral is positive because the integrand is positive on the interval of integration. Hence $f(n)>f(n+1)$ as required.

Alessio K
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  • My question is exactly why is integral of a positive function positive, rather than non-negative? I saw the proof for the latter, but not for the former (i.e. I have not seen the strict inequality case). – S11n Sep 18 '21 at 14:37
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    See https://math.stackexchange.com/questions/3026434/how-do-you-prove-the-integral-of-a-positive-function-is-also-positive/3026565 and https://math.stackexchange.com/questions/351157/is-the-riemann-integral-of-a-strictly-positive-function-positive – Alessio K Sep 18 '21 at 15:08
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    Tom still didn't use Riemann integrability up to that point, so I'm curious if there are more elementary ways to prove that. In the meantime, that answers my question. Thanks! – S11n Sep 18 '21 at 15:11
  • You're very welcome! :-) – Alessio K Sep 18 '21 at 15:12