I have to find $\int_0^\frac\pi2 \mathrm{sin}x\;dx$ by the fundamental theorem of definite integration, that is, $$\int_a^b f(x)\;dx=\lim_{n\rightarrow\infty} \sum_{r=1}^nh.f(a+rh)$$ where $h=\frac{a+b}{n}$

For the given question, $a=0$, $b=\frac\pi2$, $nh=\frac\pi2$

$$\begin{align} \int_0^\frac\pi2 \mathrm{sin}x\;dx &=\lim_{n\rightarrow\infty} h\sum_{r=1}^n\mathrm{sin}(rh)\\&=\lim_{n\rightarrow\infty} h\; (\mathrm{sin}h+\mathrm{sin}2h+\mathrm{sin}3h\cdots+\mathrm{sin}nh)\\&=\lim_{h\rightarrow0} h^2\left(\frac{\mathrm{sin}h}{h}+2\frac{\mathrm{sin}2h}{2h}+3\frac{\mathrm{sin}3h}{3h}\cdots n\frac{\mathrm{sin}nh}{nh}\right)\\&=\lim_{h\rightarrow0}h^2(1+2+3+\cdots+n)\\&=\lim_{n\rightarrow\infty}h^2\frac{n(n+1)}{2}\\&=\lim_{n\rightarrow\infty}\frac{(nh)^2}{2}\left(1+\frac1n\right)\\&=\frac{\pi^2}{8}\end{align}$$

Normal integration doesn't give this answer. So what went wrong in here?

Edit: I know how to solve this limit correctly. I just specifically wanna know what is the wrong step in this method of working. Which step or part is not allowed, or technically wrong?

  • $\begingroup$ why is $\dfrac{\sin nh}{nh} = 1?$ in fact it is $\dfrac2{\pi}$ $\endgroup$ – abel Feb 20 '15 at 2:32
  • $\begingroup$ How? I used $\lim_{\theta\rightarrow0}\frac{\mathrm{sin}\theta}{\theta}=1$ $\endgroup$ – Tejas Feb 21 '15 at 2:51
  • $\begingroup$ the error you are making is in thinking $jh$ is small and does not depend on $j.$ in my first comment i showed you that this is clearly not true for $j = n.$ $\endgroup$ – abel Feb 21 '15 at 13:25
  • $\begingroup$ Oh OK!! Now I understood. Maybe you could post it as an answer? That's exactly what I needed! :) $\endgroup$ – Tejas Feb 22 '15 at 1:49
  • $\begingroup$ i edited my answer. $\endgroup$ – abel Feb 22 '15 at 1:59

the error you are making is in thinking that all $jh$'s are small. what is happening is as $j$ increase $jh$ is not small. in fact $nh = (b-a),$ so you are not going to get $\frac{\sin nh}{nh} = 1.$

but the riemann sum $$h\sin h + h \sin 2h + \cdots h \sin jh + \cdots + h \sin nh $$ can be evaluated exactly. if you look at the characteristic triangle at the end of the rays $\theta = jh$ and $\theta = jh + h,$ you will find that $h\sin jh = \cos jh - \cos(jh + h)$ this is a telescoping series. summing it from $j = 0$ to $j = n-1$ gives you $\int_0^{\pi/2} \sin t \, dt = 1$

  • $\begingroup$ Please read the edit. $\endgroup$ – Tejas Feb 21 '15 at 2:50

The problem is that $h$ depends on $n$ and vice versa so you have to be careful when you take limits. Instead you can use the fact that $\sin x$ is the imaginary part of $e^{ix}$, so $$\sin x+\dots+\sin (nx)$$ is the imaginary part of $$e^{ix}+ e^{2ix}+\dots+ e^{inx}= \frac{1-e^{i(n+1)x}}{1-e^{ix}}-1$$


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