Cra𝓏-𝑒 coπplex

Calculus Level 5

If $y$ is the crazy function in terms of $x$ that satisfies the equation:

$\left(\frac{x\sqrt{1-y^2}}{e^\frac{\pi}{2} y^3}\right)^i-\left(\frac{y^3}{e^\frac{\pi}{2} x\sqrt{1-y^2}}\right)^i=2y$

Then compute:

$\int_{-\infty}^{0}1+y(x) \hspace{2px} dx+\int_0^{\infty}1-y(x) \hspace{2px} dx$

The answer can be written in the form $\hspace{1px}\cfrac{ae^{\frac{\pi}{a}}+ae^{-\frac{\pi}{a}}}{b}$ , where $a$ and $b$ are coprime integers. Enter $|a|+|b|$ .

Notes: $\hspace{3px}i=\sqrt{-1}$

Assume the positive branch of the square root is used.

Hint : If you end up using any inverse or trigonometric functions, use the main branch, for example: $\hspace{2px}\arcsin r=\theta+2k\pi$ with $k=0$

The answer is 7.

1 solution

Saúl Huerta
May 14, 2021

The first step is to transform the equation into something more manageable to check if there even exists an expression for $y$ in terms of known functions. We can see that the $e^\pi$ can be factored out of both terms to get the constant $e^{-i\frac{\pi}{2}}$ , which is equal to $-i$ , which is equal to $\frac{1}{i}$ . We can divide both sides by $2$ and use the identity $\frac{e^{i\theta}-e^{-i\theta}}{2i}=\sin\theta$ to rewrite as follows: $y=\frac{\left(\sqrt{\frac{x^2(1-y^2)}{y^6}}\right)^i-\left(\sqrt{\frac{x^2(1-y^2)}{y^6}}\right)^{-i}}{2i}=\sin\left(\ln\sqrt{\frac{x^2(1-y^2)}{y^6}}\right)$ $\implies e^{\arcsin y}=\frac{x\sqrt{1-y^2}}{y^3}$ It is clear that $y(x)\in(-1, 1)$ and the function is continuous for $x\in\mathbb{R}$ . Now, finding an expression for $y$ seems to be very difficult, if not impossible. Instead let's take a look at the graph of its inverse function $y^{-1}(x)=\frac{x^3e^{\arcsin x}}{\sqrt{1-x^2}}$ : Since the inverse function is obtained by reflecting the function with respect to the $x$ -axis and then rotating the original function 90 degrees the integral we are trying to compute becomes: $\int_{-1}^{1}\frac{x^3e^{\arcsin x}}{\sqrt{1-x^2}}\hspace{2px} dx$ In its current form, this integral is screaming for a trig sub, so we substitute $\hspace{2px} x=\sin\theta$ : $=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^\theta \sin^3\theta \hspace{2px} d\theta$ Here we could do some integration by parts but I find it much better if we do some neat substitutions to greatly simplify the calculations. We know that $\hspace{1px}\sin^3\theta=\frac{3\sin\theta-\sin3\theta}{4}$ and $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ . After substituting these identities in succession respectively and simplifying we end up with four very simple, neat and easy exponential integrals: $=\frac{3}{8i}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{(1+i)\theta}\hspace{2px} d\theta-\frac{3}{8i}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{(1-i)\theta}\hspace{2px} d\theta-\frac{1}{8i}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{(1+3i)\theta}\hspace{2px} d\theta+\frac{1}{8i}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{(1-3i)\theta}\hspace{2px} d\theta$ Integrating and doing all the algebra yields: $\int_{-1}^{1}\frac{x^3e^{\arcsin x}}{\sqrt{1-x^2}}\hspace{2px} dx=\frac{2e^{\frac{\pi}{2}}+2e^{-\frac{\pi}{2}}}{5}$ Therefore $a+b=2+5=\boxed{7}$

@Saúl Huerta That was a fun problem! Even though I did the same as you to solve it, I'm still not sure about some details where you "invert" the functional equation: $\begin{aligned} y &= \sin\left(\ln\sqrt{\frac{x^2(1-y^2)}{y^6}}\right) =: f(x,y), &&& f(-x,y)&=f(x,y), &&& |y|&=|f(x,y)|\leq 1 \end{aligned}$

Shouldn't $y(x)$ be mirror-symmetric to the y-axis? If $(x,y)$ is a solution, so is $(-x,y)$ after all (closely related to 3.)
The functional equation has multiple solutions: $\begin{aligned} y&=f(x,y)&&&\Rightarrow &&&& \ln\sqrt{\frac{x^2(1-y^2)}{y^6}}&=\arcsin(y)+2k\pi &&&\vee &&&& \ln\sqrt{\frac{x^2(1-y^2)}{y^6}}&=-\arcsin(y)+(2k+1)\pi &&&\Bigl|k&\in\mathbb{Z} \end{aligned}$ These additional constants become factors when we solve for $x$ and thereby change the value of the integral later
What happened to the absolute values when we simplify the radical? Shouldn't the fundamental solution for $k=0$ be $\red{|x|}=\frac{\red{|y|}^3}{\sqrt{1-y^2}}e^{\arcsin(y)}$ As far as I can tell, the functional equation remains valid for arbitrary choices of signs for $x$ , so one should be able to construct a discontinuous function $y(x)$ that satisfies the functional equation. Of course such a function might not be invertible anymore, e.g. $\begin{aligned} g : &&(-1;\:1)&\rightarrow\mathbb{R}, &&& g(y)&:=\frac{y^3}{\sqrt{1-y^2}}e^{\arcsin(y)}\quad\text{invertible}, &&&&&&y(x)&:=\begin{cases} g^{-1}(-x):&x\in(-1;\:1)\cap \mathbb{Q}\\ g^{-1}(x):&\text{else} \end{cases} \end{aligned}$ As far as I can tell, $y(x)$ satisfies the functional equation, is discontinuous (almost) everywhere and is not invertible. The integrals would not converge for this choice of $y(x)$ , because it has too many discontinuities.

Suggestion: Define $y$ as a continuous function with $\operatorname{sgn}(x)=\operatorname{sgn}(y)$ (or something similar) to get rid of the problems with absolute values. Not sure what to do about the set of solutions $k\in\mathbb{Z}$ ...

Carsten Meyer - 3 weeks, 5 days ago

@Carsten Meyer Hey thanks for your very detailed reply! Sorry if the problem was vague but I tend to forget some important technical nuances which is what happened here. With respect to the absolute value, I will simplify the radical within the functional equation so that there are no discontinuities (which is why I originally intended but when writing the problem in the latex editor I decided to put the whole fraction in the radical for aesthetic purposes, but now that I think about it it was stupid lol): $\left(\frac{x\sqrt{1-y^2}}{e^\frac{\pi}{2} y^3}\right)^i-\left(\frac{y^3}{e^\frac{\pi}{2} x\sqrt{1-y^2}}\right)^i=2y$

And about the cyclic properties of the trigonometric function within the problem I will specify to use the main branch $k=0$ . Thanks for helping me remember the theory and maintain mathematical consistency! I sometimes forget it lol, will edit the problem asap :D. (Edit: I edited the problem I think now it should be fine, if there is still something wrong I would be very glad if you let me know. In the future maybe I will even leave the multiple solutions for $k\in\mathbb{Z}$ including the branch with the negative inverse sine, but for now I will leave the specifications as they are while I work in the more general solution. The absolute value problems do have to go though to ensure continuity and to allow for an inverse function).

Saúl Huerta - 3 weeks, 5 days ago

@Saúl Huerta You're welcome! I know I'm sort of nitpicky about these details - it's the way some people learn ;) Mentioning the fundamental branch of square-root and the inverse trig functions should do the trick!

Carsten Meyer - 3 weeks, 5 days ago

Cra𝓏-𝑒 coπplex

The answer is 7.

1 solution

0 pending reports