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prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$

I showed that in $$ \forall x \in [0,\frac{\pi}{4}] \quad \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$ but it's not good enough

3SAT
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Daniel Katzan
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4 Answers4

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Using the fact that $\sin(x)\le x$ for $x\ge0$, $$ \begin{align} \int_0^{\pi/4}\frac{1-\cos(x)}x\,\mathrm{d}x &=\int_0^{\pi/4}\frac{2\sin^2(x/2)}x\,\mathrm{d}x\\ &=\int_0^{\pi/4}\frac{\sin(x/2)}{x/2}\cdot\sin(x/2)\,\mathrm{d}x\\ &\le\int_0^{\pi/4}1\cdot\frac x2\,\mathrm{d}x\\[4pt] &=\frac{\pi^2}{64} \end{align} $$

robjohn
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Another way forward is to recall the Taylor series for the cosine

$$\cos x =\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \tag 1$$

Noting that for $0\le x\le \pi/4$ the terms of the series in $(1)$ alternate signs and are decreasing monotonically. Therefore, we have

$$\cos x \ge 1-\frac12 x^2$$

so that

$$\frac{1-\cos x}{x}\le \frac12 x$$

Finally, we see that

$$\begin{align} \int_0^{\pi/4}\frac{1-\cos x}{x}\,dx &\le \int_0^{\pi/4} \frac12 x\, dx\\\\ &=\frac{\pi^2}{64} \end{align}$$

as was to be shown!

Mark Viola
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  • You might want to show that $1-\cos(x)-\frac{x^2}2$ is concave and its derivative and value at $x=0$ is $0$. This shows that $1-\cos(x)\le\frac{x^2}2$. This is a bit cleaner than worrying about the trailing part of the series. – robjohn Aug 06 '15 at 20:29
  • @robjon Well, yes, but the trailing part alternates sign so the inequality holds for $x$ in the interval of interest. – Mark Viola Aug 06 '15 at 20:39
  • And the terms are monotonically decreasing. Just one more thing you need to state and check when truncating series. – robjohn Aug 06 '15 at 20:47
  • @robjon All evident for the excrutiatingly well-known series for the cosine, don't you agree. – Mark Viola Aug 06 '15 at 20:51
  • However, here we are using $$\cos(x) = 1-\frac12x^2+\frac1{24}x^4 +O!\left(x^6\right)$$ to derive $$\frac{1-\cos(x)}x\le\frac12x$$ and without some backstage help, the floor show is a bit confusing. It is okay as a way to point the OP, however. – robjohn Aug 06 '15 at 21:11
  • @robjohn I had been working from a "smart phone" and am back on the laptop. I can better understand your perspective now and agree that for some users, the development could be obfuscating. So, stepping back, I have rewritten a bit and hope that this proves more useful for a broader range of possible "readers." And I know that you need help here too since you're a novice at math ... LOL. – Mark Viola Aug 06 '15 at 21:47
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Hint: $1-\cos(x)=2\sin^2\left(\frac{x}{2}\right)$ and $0 \leq \frac{\sin(t)}{t} < 1$ for all $t \in \left(0,\frac{\pi}{2}\right]$. If done properly, you should get $\int\limits_0^{\pi/4}\,\frac{1-\cos(x)}{x}\,\text{d}x<\frac{\pi^2}{64}$ (which I think you meant to do, since $\frac{\pi^2}{4}$ is a very weak bound).

Batominovski
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Since: $$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right) $$ over the interval $\left[0,\frac{\pi}{4}\right]$ we have: $$ \frac{1-\cos x}{x}=\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{x}{2}\leq\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2} $$ so: $$ \int_{0}^{\frac{\pi}{4}}\frac{1-\cos x}{x}\,dx \color{red}{\leq} \int_{0}^{\pi/4}\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2}\,dx = \frac{\pi^2}{64}-\frac{191\,\pi^2}{3\cdot 2^{18}}.$$

Jack D'Aurizio
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