2
$\begingroup$

The Question i have is: Calculate the following Riemann Integral
$$\int_0^\frac{\pi}3 \tan(x) \,dx.$$

I know that $\int_a^b f(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta X$
and so I've worked out $\Delta X = \frac {b-a} n = \frac {\frac \pi 3} n = \frac \pi {3n}$
and also $ x_i^* = a+ (\Delta X)i = 0 + (\frac \pi {3n})i$.

So for my question I know that the $\int_0^ \frac\pi 3 tan(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n \tan((\frac \pi {3n})i) \times \frac \pi {3n} $

but I am not 100% sure where to go from here.

  • 0
    Not sure if this is the best way to go about computing this integral. You really want to consider the antiderivative of $\tan{x}$ and use the Fundamental Theorem of Calculus.2012-12-23
  • 1
    When I look at this (and the answer using the FToC) - and especially the result - I asked myself if one could compare this series to the alternating harmonic series. For example bound it from above or below by the alternating series. I don't know if this could work, but it would seem interesting. At least the way using the FToC is far more efficient and usable.2012-12-23
  • 0
    @AndreasS Looking at a [picture](http://i.stack.imgur.com/SQXdk.png) suggests that the tangent curve is the same as the $1/x$ curve flipped horizontally at an axis near $1$ and slightly shifted upwards. Perhaps one can figure out how to center the point of intersection of both functions in $[0,\pi/3]$ so that one can compute the integral of $\tilde{f}(x)$ obtained from $1/x$ instead of $\tan (x)$.2012-12-25

2 Answers 2

2

Calculating integrals this way can be very hard... That's why we have the Funcdumental Theorem of Calculus:

$$\int \tan x\, dx=\int\frac{\sin x}{\cos x}\, dx$$ Substituting $u=\cos x $ yields $$\int \tan x\, dx=\int\frac{-1}{u}\, dx=-\ln\left|u\right|+c=-\ln\left|\cos x\right|+c$$ Therefore, by the 2nd Fundumental Theorem of calculus, $$\int_0^\frac{\pi}3 \tan(x)\, dx=-\ln\frac12+\ln1=\ln 2$$ This also implies that $$\lim_{n\to\infty} \sum_{i=1}^n\frac{\pi}{3n} \tan((\frac \pi {3n})i)=\ln 2 $$

  • 1
    @rlgordonma i fully understand the above method in order to work out integrals as it was one of the first ways i was taught, however the question im currently doing asks for me to work the above integration using Riemanns Integral2012-12-23
  • 0
    @jill The integral I am using is the Riemann Integral2012-12-23
  • 0
    i understand the approach you took, however, that doesn't explain how you've worked it out using riemann's integral? as you've used th fundamental theorem and just in putted the answer into the riemanns summation2012-12-23
  • 0
    The indefinite integral is not the Riemann Integral. The definite one is however, and is related to the indefinite one via the 2nd fundumental theorem of calculus2012-12-23
  • 0
    ohhhhh okay now i see, so to compute this simply using riemanns theory would not be possible? as we'd have to use the rules of fundamental calculus?2012-12-23
  • 0
    @jill I don't know if it's possible, but it would certainly be very complicated2012-12-23
  • 0
    okayyy :) well im a stage1 maths student so im assuming this is the method required to answer our question, thankyou :)2012-12-23
  • 0
    @jill - do you mean evaluating various finite versions of that sum to approximate the integral, knowing that the answers must be increasingly close to $\log{2}$? Then that is what you must do - you can use Matlab or Mathematica or some script to automate this. I do not know a closed form for the sum, so I cannot imagine how that would help.2012-12-23
  • 0
    @rlgordonma im assuming the question isn't asking for the evaluation as it simple asks "calculate the following Riemann integrals" and then gives a couple of questions2012-12-23
0

The integral in question can indeed be computed as a limit of Riemann sums.

We consider Riemann sums $$R_N:=\sum_{k=1}^N \tan(\xi_k)(x_k-x_{k-1})\qquad(1)$$ where the partition $0=x_0

Fix $k$ for the moment. Then $$x_k-x_{k-1}=\arccos'(\tau)\bigl(2^{-k/N}-2^{-(k-1)/N}\bigr)\qquad(2)$$ for some $\tau\in\bigl[2^{-k/N},\>2^{-(k-1)/N}\bigr]$. Now

$$\arccos'(\tau)={1\over\cos'(\arccos\tau)}=-{1\over \sin\xi}\ ,\qquad(3)$$ where $\cos\xi=\tau$. It follows that $$2^{(k-1)/N}\leq{1\over\cos\xi}\leq 2^{k/N}$$ or $${1\over\cos\xi}=2^{k/N}\cdot 2^{-\Theta/N}$$ for some $\Theta\in[0,1]$. Now chose this $\xi$ as the $\xi_k$ in $(1)$. Then we get, using $(2)$ and $(3)$: $$R_N=\sum_{k=1}^N{\sin\xi_k\over \cos\xi_k}{1\over\sin\xi_k}\bigl(2^{-(k-1)/N}-2^{-k/N}\bigr)=\sum_{k=1}^N 2^{-\Theta_k/N}(2^{1/N}-1)\ .$$ For large $N$ the factors $2^{-\Theta_k/N}$ are arbitrarily close to $1$. Therefore the last sum essentially consists of $N$ terms of equal size $2^{1/N}-1$. (The obvious squeezing argument can be supplied by the reader.) It follows that $$\lim_{N\to\infty} R_N=\lim_{N\to\infty}{2^{1/N}-1\over 1/N}=\log 2\ .$$