Додаток: інтегрування з використанням формули Ейлера

Euler's famous identity states that: $$e^{ix} = \cos x + i\sin x$$ Replacing $x$ with $-x$, we get: $$e^{-ix} = \cos x - i\sin x$$ We can then combine these $2$ equations to obtain the following: $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$ $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ The key here is that the trigonometric functions can be written in terms of exponential functions. When you combine trigonometric functions, things can often get very messy, but exponentials are generally far less messy when put together.

One can also make use of the real and imaginary parts of a complex number: $$\cos x = Re(e^{ix})$$ $$\sin x = Im(e^{ix})$$

It may seem ironic, but introducing complex numbers can actually make many integration and summation problems simpler.

Let us apply the identity $\cos x = Re(e^{ix})$: $$\int e^x\cos x dx = \int e^x Re({e^{ix}})dx$$ $$\int e^x\cos x dx = Re \int e^x e^{ix}dx$$ $$\int e^x\cos x dx = Re \int e^{(i+1)x}dx$$ $$\int e^x\cos x dx = Re\left[\frac{e^{(1+i)x}}{1+i}\right]+C$$ $$\int e^x\cos x dx = e^x Re\left[\frac{e^{ix}(1-i)}{2}\right]+C$$ $$\int e^x\cos x dx = e^x\frac{\cos x + \sin x}{2}+C$$

We have successfully computed the integral without any integration by parts!

Let's apply the identiy $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$: $$\int \sin^2 x dx = \int \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2 dx$$ $$\int \sin^2 x dx = \int \frac{e^{2ix}-2+e^{-2ix}}{-4} dx$$ $$\int \sin^2 x dx = \frac{x}{2} + \frac{-e^{2ix}+e^{-2ix}}{8i} + C$$ $$\int \sin^2 x dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C$$ Normally, we would use the double-angle identity to reduce the power of the trigonometric functions, but this method is quite painless. In this particular simple example, there is no real benefit to using this method, but evaluating more complicated integrals may bring to light the utility of these identities.

Submitted by Sujay Kazi

Define $f(x)$ as follows: $$f(x) = \sum_{n=0}^{\infty}\frac{\sin(nx)}{n!}.$$ Evaluate $f\left(\frac{\pi}{3}\right)$.

(This is not actually an integration, but it is a very nice problem that employs the same principles taught in the above lesson. Also, the answer is too complicated to type into an answer box nicely, so just write down your answer and look at the solution when you are ready.)

Source: Stanford Math Tournament 2013: Calculus Tiebreaker, #3

[Click Continue for the first hint.]

Hint 1

Use the fact that $\sin(nx) = Im(e^{inx})$.

[Click Continue for the second hint.]

Hint 2

Use the following Maclaurin series to simplify the summation: $$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

[Click Continue for the solution.]

We use the fact that $\sin(nx) = Im(e^{inx})$ and that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$. It follows that

$$\begin{align}f(x) = \sum_{n=0}^{\infty}\frac{\sin(nx)}{n!} &= Im\sum_{n=0}^{\infty}\frac{(e^{ix})^n}{n!} \\&= Im\left(e^{e^{ix}}\right) \\&= Im\left(e^{\cos x + i\sin x}\right)\\& = e^{\cos x}\sin(\sin x)\end{align}$$

Plugging in $x = \frac{\pi}{3}$ gives us our answer: $$f\left(\frac{\pi}{3}\right)=e^{1/2}\sin\left(\frac{\sqrt{3}}{2}\right).$$