-3
$\begingroup$

Using the following definition of $\sin(x)$

$$\sin(x) \stackrel{\text{def}}{=} \frac{1}{2}\left(e^{ix} - e^{-ix}\right)$$

Results in the following integral

$$\begin{align} \int \sin(x)\ dx &= \frac{1}{2}\int\left(e^{ix} - e^{-ix}\right) \ dx \\ &= \frac{1}{2i}\left(e^{ix} + e^{-ix}\right) + C \end{align}$$

But $\int \sin(x)\ dx = -\cos(x) + C \iff \int \sin(x)\ dx = \frac{1}{2}\int\left(e^{ix} + e^{-ix}\right) + C$. Thus the $\frac{1}{i}$ multiplicand is the term here is what is producing the wrong integral.

Is it only possible to integrate $\sin(x)$ and prove $\int \sin(x) = -\cos(x) + C$ via the Taylor Series definition of $\sin(x)$?

$$\sin(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} \ \ \ \ \ \text{(Taylor Series Definition)}$$

$\endgroup$
  • 6
    $\begingroup$ Your starting definition is missing $i$ at the denominator. $\endgroup$ – Yves Daoust Aug 8 '16 at 9:30
  • 3
    $\begingroup$ Hint: Your definition of the sine function is wrong, so is your antiderivative of $\sin (x)$. $\endgroup$ – Nigel Overmars Aug 8 '16 at 9:30
  • 3
    $\begingroup$ Also, $\int\sin(x)dx$ is not $\cos(x) + C$... $\endgroup$ – 5xum Aug 8 '16 at 9:43
3
$\begingroup$

You made the usual mistake in mathematics: sloppyness!

Using the following definition of $\sin(x)$

$$\sin(x) \stackrel{\text{def}}{=} \frac{1}{2}\left(e^{ix} - e^{-ix}\right)$$

Wrong. The formula you wrote evaluates to

$$\frac12(\cos x + i\sin x - (\cos(-x) + i\sin(-x))) = \frac12(\cos x + i\sin x - \cos x + i\sin x) =\frac12 (2i\sin x) = i\sin x \neq \sin x$$


Using the correct formula for $\sin x$ (which is $\frac1{2i}(e^{ix}-e^{-ix}$) will get you:

$$\int \sin x dx = \frac{1}{2i}\int(e^{ix}-e^{-ix})dx =\\ =\frac{1}{2i}\left(\int e^{ix} dx - \int e^{-ix} dx \right)=\\ =\frac{1}{2i}\left(\frac{1}{i}e^{ix} - \frac{1}{-i} e^{-ix}\right)+C=\\ =\frac{1}{2i}\cdot \frac{1}{i}\left(e^{ix} + e^{-ix}\right)+C=\\ =\frac{1}{-2}(\cos x + i\sin x + \cos(-x) + i\sin(-x))+C=\\ =\frac{1}{-2}(\cos x + i\sin x + \cos(x) - i\sin(x))=\\ =\frac{1}{-2}\cdot2\cdot \cos x+C = -\cos x+C$$

Which works out to what you would expect.

$\endgroup$
  • 1
    $\begingroup$ I like "You made the usual mistake in mathematics: sloppyness!" :) +1 $\endgroup$ – 6005 Aug 8 '16 at 10:11
3
$\begingroup$

Notice, Euler's formula (and de Moivre formula):

$$e^{\theta i}=\cos(\theta)+\sin(\theta)i$$

And use:

  • $$\cos(-\theta)=\cos(\theta)$$
  • $$\sin(-\theta)=-\sin(\theta)$$

So, we get:

$$e^{\theta i}-e^{-\theta i}=\left(\cos(\theta)+\sin(\theta)i\right)-\left(\cos(-\theta)+\sin(-\theta)i\right)=$$ $$\cos(\theta)+\sin(\theta)i-\cos(\theta)+\sin(\theta)i=2\sin(\theta)i$$

So:

$$\sin(\theta)=\frac{e^{\theta i}-e^{-\theta i}}{2i}$$

Now, the integral become:

$$\int\sin(\theta)\space\text{d}\theta=\int\frac{e^{\theta i}-e^{-\theta i}}{2i}\space\text{d}\theta=\frac{1}{2i}\left[\int e^{\theta i}\space\text{d}\theta-\int e^{-\theta i}\space\text{d}\theta\right]=$$ $$\frac{-ie^{\theta i}-ie^{-\theta i}}{2i}+\text{C}=\text{C}-\cos(\theta)$$

$\endgroup$
1
$\begingroup$

By the Euler and de Moivre formulas,

$$e^{ix}-e^{-ix}=(\cos x+i\sin x)-(\cos x-i\sin x)=2i\sin x.$$


Other check:

$$(e^{ix}-e^{-ix})^2=e^{2ix}-2+e^{-i2x}=2\cos(2x)-2,$$ which is a negative number !


The developments of $e^{ix}$ and $e^{-ix}$ differ in the sign of the terms of odd power, so that when you subtract them, only the odd powers remain, and $i^{2k+1}=\pm i$.


You can establish two integrals in a single go, as follows:

$$e^{ix}=\cos x+i\sin x,$$ then omitting the constant,

$$\int e^{ix}dx=\frac{e^{ix}}i=\sin x-i\cos x.$$ Then equate the real and imaginary parts.

$\endgroup$
  • 1
    $\begingroup$ $2 \cos (2x)-2$ isn't always negative, it is certainly always non-positive though :-) $\endgroup$ – Kevin Aug 8 '16 at 9:47
  • 1
    $\begingroup$ @Bacon: with $x$ uniformly spread in $[0,2\pi)$, the expression is almost certainly negative. :) $\endgroup$ – Yves Daoust Aug 8 '16 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.