A calculus problem by Refaat M. Sayed

Calculus Level 5

$\int \limits^{\infty }_{0}\frac{\cos \left( \pi x\right) }{4x^{2}-1} dx=\frac{-\pi }{A}$ Find $A$ .

This problem similar to problem posted by Ariel Gershon

The answer is 4.

1 solution

Guillermo Templado
Nov 24, 2016

Applying residue theorem of complex analysis.

The function $f(x) = \frac{e^{i\pi x}}{4x^2 -1}$ , have isolated singularities at the roots of $4x^2 - 1 = 0$ what are $x_1 = 1/2, \space x_2 = - 1/2$

$\displaystyle \left(\text{Res f(x)}; \space x = x_1 = 1/2 \right) = \lim_{x \to 1/2} \frac{(x - 1/2) \cdot e^{i\pi x}}{4x^2 -1} = \frac{i}{4}$ $\displaystyle \left(\text{Res f(x)}; \space x = x_1 = -1/2 \right) = \lim_{x \to -1/2} \frac{(x + 1/2) \cdot e^{i\pi x}}{4x^2 -1} = \frac{i}{4}$

Note that $\int_{\infty}^{-\infty} \frac{\cos \pi x}{4x^2 -1} \, dx = - \int_{- \infty}^{\infty} \frac{\cos \pi x}{4x^2 -1} \, dx$ Therefore, applying residue theorem (Below, in the comments section, is the residue theorem and the test is more detailed) $2 \cdot \int_{- \infty}^{\infty} \frac{e^{i\pi x}}{4x^2 -1} \, dx = 2i\pi \cdot \left( \frac{2i}{4} \right) = -\pi \Rightarrow \int_{- \infty}^{\infty} \frac{e^{i\pi x}}{4x^2 -1} \, dx = \frac{-\pi}{2}.$ $\displaystyle \int_{-\infty}^{\infty} \frac{\cos(\pi x)}{4x^2 - 1} \,dx = 2 \left(\int_{0}^{\infty} \frac{\cos(\pi x)}{4x^2 - 1} \, dx \right) \Rightarrow \int_{0}^{\infty} \frac{\cos(\pi x)}{4x^2 - 1} \, dx$ is the half(middle) of real part of $\int_{- \infty}^{\infty} \frac{e^{i\pi x}}{4x^2 -1} \, dx = \frac{-\pi}{2} \Rightarrow \int_{0}^{\infty} \frac{\cos(\pi x)}{4x^2 - 1} \, dx = \frac{-\pi}{4}$

Hi @Pi Han Goh and @Guillermo Templado at first sorry for late reply. Second I will go for pi question. My solution is depends on Residue theorem. . To solve this integral we will use Cauchy principal value $\text{PV}\int \limits^{\infty }_{-\infty }\text{f}\left( x\right) \cos \left( mx\right) dx = \Re \{ 2\pi i\left( \sum {R} +\frac {1}{2} \sum {\grave{R} }\right) \}$ Where $\sum {R}$ is thethe sum of the residues of $f (x)e^{imx}$ at all poles lying in the upper half plane not including those on the real axis and $\sum {\grave{R}}$ isthe sum of the residues of $f(x)e^{imx}$ at all simple poles lying on the real axis.so the integral will be $\frac{\Re \{ 2\pi i\left( 0+\frac{\frac{i}{4} +\frac{i}{4} }{2} \right) \} }{2} =\frac{-\pi }{4}$ . Now I will go for Guillermo. Your solution contains one mistake $\frac{\int \limits^{\infty }_{-\infty }\text{Even function }d (\text{variable })}{2} =\int \limits^{\infty }_{0}\text{Even function }d(\text{variable })$ and don't forget $\Huge{dx}$ in your integrals

Refaat M. Sayed - 4 years, 6 months ago

Ammended $\Huge {dx}$ . Now I have a question, don't forget what is $\Huge{f(x)}$ ? and why you divide by 2 in your last point of your solution....? And what is 5 in your first line? And this corollary of residue theorem is applied for what functions ?...

Guillermo Templado - 4 years, 6 months ago

For the first question : $f (x)$ is an integrable function at any point in the interval and finite at any point except at poles the function will be infinity.

For the second : $\int \limits^{\infty }_{-\infty }\text{Even function }\text{d}\left[ \text{variable }\right] =\overbrace{\int \limits^{0}_{-\infty }\text{Even function d}\left[ \text{variable }\right] } \limits^{I_{1}}+\overbrace{\int \limits^{\infty }_{0}\text{Even function d}\left[ \text{variable }\right] } \limits^{I_{2}}$
And we have $I_{1}=I_{2}$ .

For the third question : it's $\frac {1}{2}$ not $5$

For the fourth question : this corollary applied for functions that has real roots make the function $\to \infty$ .
BTW you forget $dx$ in the last integral $\ddot\smile$

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – First. What is $\Huge{f(x)}$ ?.

Second?.I didn't have any mistake with the even function (your second point), because I wrote it good.

Third?Now it's right, but then,

Fourth? what happen whith your fourth point? Please, can you write the corollary or repeat what are you saying?

And finally, yes I forgot $dx$ ...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – I have answered all your questions before and I did my best. For your fourth question I will post some problems to show the difference between your solution and mine( I have said this difference at my fourth point).

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – ok, I agree... No problem... but if you mean $f(x) = \frac{1}{4x^2 - 1}$ your proof doesn't convince me... because $q(x) = 4x^2 - 1$ have their roots at the real axis...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – Yes, you are right. This function has two real roots make the function equal infinity. And that is the reason we use Cauchy principal value.

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – I've ammended my mistake with the even function. Are you sure, don't you have any doubt here, in my proof? Please , be sincere... Use wolfram alpha, what is? $\displaystyle \int_{0}^3 \frac{\cos \pi x}{4x^2 - 1} \, dx$ $\displaystyle \int_{0}^4 \frac{\cos \pi x}{4x^2 - 1} \, dx$ $\displaystyle \int_{0}^5 \frac{\cos \pi x}{4x^2 - 1} \, dx$ And what is the winding number(number of loops,turns) of the path $\gamma(x) = e^{2i \pi x}; \space x \in [0,1]$ , and what is the winding number(number of loops,turns) of the path $\gamma(x) = e^{i \pi x}; \space x \in [0,1]$ ? what happens here? why can't I apply my 2 propositions used in the other problem?

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – We can't apply the second proposition in the other problem simply because we have in this problem real roots not imaginary . For the question the winding number for the path $\gamma(x )=e^{2i\pi}$ is $1$ and for the second function is $-1$ . It's clear now, or I miss something ?

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – Your first part is exact, good... For the question the winding number(number of "loops","turns"), the windinding number "measure the speed of these paths for giving a turn around 0 in a time of $2\pi$ ". Tomorrow, I'll write a formal definition of the winding number and residue theorem and other propositions...For example, $\gamma (x) = e^{2i \pi x}; \space x \in [0, 1]$ the winding number is 1,i.e, this path is the circumference of radius 1 with center 0 to the "speed of 1 giving a loop around 0". This implies that $\gamma_1 (x) = e^{i \pi x}; \space x \in [0, 1]$ is the semicircurference (in the uper-half plane) of radius 1 with center 0, has a "speed" $\frac{1}{2}$ of $\gamma (x) = e^{2i \pi x}; \space x \in [0, 1]$ , then its winding number is... Tomorrow, I'll continue... Furthemore, you have to have on mind... ok, tomorrow... now I can't... I told you I had a personal life... good, right now I'm going to write music... I'm really, really bussy...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – Ok, first of all, there is a long way so far... In Spain, we say: there is a long way to play "hard" rock, but there is a longer way to play Flamenco .

Definition.- If $\gamma : [0,1] \to \mathbb{C}$ is a closed piecewiese smooth path and $a \notin \gamma([0,1])$ , the winding number of $\gamma$ respect to $a$ (number of turns around $a$ by $\gamma$ is: $\displaystyle \text{ Ind(}\gamma , a) = \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{\gamma '(s)}{\gamma(s) - a} \, ds = \frac{1}{2i \pi} \oint_{\gamma} \frac{1}{z - a} \, dz$

Example.-

$\gamma (x) = e^{2i \pi x}; \space x \in [0, 1]$ the winding number around $0$ is $\displaystyle \text{ Ind(}\gamma , 0) = \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{2i \pi \cdot e^{2i \pi x}}{e^{2i \pi x}} \, dx = 1$

To be continued..

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – Then,its, clear that if $\gamma_1 (x) = e^{i \pi x}; \space x \in [0, 1]$ , $\displaystyle \frac{1}{2i \pi} \cdot \int_{0}^{1} \frac{i \pi \cdot e^{i \pi x}}{e^{i \pi x}} \, dx = \frac{1}{2}$ . Nevertheless, we can't say $\displaystyle \text{ Ind(}\gamma_1 , 0) = \frac{1}{2}$ because $\gamma_1$ is not closed . Sorry, I want to make this quickly and right now, but it's impossible for me, I have to go slowly... To be continued... BTW, what means closed? do you understand me so far, friend? (I'm doind this for you)...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – That is very helpful I understand what you write my friend, thanks.

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – thanks for you friend, I'll continue with other generalization of this definition and later 3 or 4 propositions, the residue theorem, and I'll explain what I did in my first part of my proof, and the end... To be continued... I promise in two days at most, this is finished. Hold on, tomorrow I hope to work very hard here, trying to explain "almost everything". Greetings for you and Pi, Pi can follow our coversation, ask, etc... work here ,haha, etc etc..

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – Thanks for the best coversation i had in brilliant till now.. I will wait your analysis

Refaat M. Sayed - 4 years, 6 months ago

@Refaat M. Sayed – good, my idea final is $\displaystyle \lim_{x \to 1/2} \frac{\cos (\pi x)}{4x^2 - 1} = \lim_{x \to -1/2} \frac{\cos (\pi x)}{4x^2 - 1} = \frac{-\pi}{4}$ ,(for example, you can use L'Hopital), i,e, the isolated singularites are avoidable. Ok, I'm going to walk for a walk with my dogs and I keep on with "my analysis"... Today, there is a long way for me,I'm thinking maybe I'm wrong...

The winding number of a path $\gamma$ respect to $a$ is constant in the connected component containing (or not) $a$ , This is, the winding number is 0 if $\gamma$ doesn't turn around $a$ and it's constant if $\gamma$ turns around $a$ . (Think about the winding number as the number of turns around...)

Proposition (generalization of the previous definition) .-

If $C_r (t) = a + re^{it}; t \in [0, 2\pi]$ then $\text{ Ind(C}_r, z) = 1$ if $|z - a| < r$ and $\text{ Ind(C}_r, z) = 0$ if $|z - a| > r$

Proof.-

if $|z - a| < r$ then $\text{ Ind(C}_r, z) = \frac{1}{2i \pi} \cdot \int_{0}^{2 \pi} \frac{ir e^{it}}{re^{it}} \,dr = \frac{1}{2i \pi} \oint_{C_r} \frac{1}{w - a} \,dw =1$

if $|z - a| > r$ then $\text{ Ind(C}_r, z) = 0$ due to previous argument $\square$

Definition

If $\gamma_i, i =1, ... , m$ are closed paths, the union(direct sum) of these paths is called a cycle $\Gamma$ and $\displaystyle \text{ Image }(\Gamma) = \cup_{j = 1}^{m} \text{ Image } (\gamma_j)$

to be continued

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – I haved added a previous definition and the calculation of the above limits... If you try to calculate $\int_{0}^{\infty} \frac{\cos mx}{4x^2 - 1} \, dx$ you'll see that this integral only converges if $m = \pi$ if $m \in [0, 2\pi]$ (You can use wolfram alpha)

Definition

Two cycles $\Gamma, \Delta$ in the open set $\Omega \subseteq \mathbb{C}$ are $\Omega$ - homologue if $\displaystyle \text{ Ind (}\Gamma , z) = \sum_{j = 1}^m \text{ Ind (}\gamma_j , z) = \text{ Ind (}\Delta , z), \space \forall z \notin \Omega$ . A cycle $\Gamma \subset \Omega$ is $\Omega$ - homologue to $0$ if $\text{ Ind (}\Gamma , z) = 0, \space \forall z \notin \Omega$ .

Now, I'm going to give (maybe) the most important theorem in complex analysis, but I need the definition of harmonic function... It's not necessary, but I want to do it...

Sorry Pi and Refaat, to be continued... How many "to be continued" there will be,haha? I think, you finish killing me...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – Relevant wiki Cauchy integral formula .

I'm thinking in other possible solution... (Later, I'll see if I can use it... when I finish here there is other solution from other person)

Proposition (Homologous version of Cauchy's integral formula)

If $\Gamma$ is a piecewise smooth cycle in the open set $\Omega \subset \mathbb{C}$ , $\Omega$ - homologue to $0$ , then $\forall f \in \mathcal{H}(\Omega)$ and every $z \in \Omega \text{ - Image(}\Gamma)$ : $\displaystyle \text{ Ind (}\Gamma , z) \cdot f(z) = \frac{1}{2i \pi} \oint_{\Gamma} \frac{f(w)}{w - z} \, dw$ ... To be continued, sorry. In one hour, I'll keep on...

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – This part is not necessary... Sorry for my English and names...

Definitions.-

a) $\mathbb{C}_{\infty} = \mathbb{C} \cup \{\infty\}$ is compacting of Alexandroff de $\mathbb{C}$

b) Let $\Omega \subseteq \mathbb{R}^2$ be an open set in $\mathbb{R}^2$ and $u : \Omega \to \mathbb{R}$ be a function such that $u \in \mathcal{C}^2 (\Omega)$ , then the Laplacian operator for $u$ is $\Delta u = \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}$ . Then $u$ is an harmonic funcion if $\Delta u = 0$ .

Now, I'm going to write the fundamental theorem of complex Analysis in a variable for me.

Theorem.-

If $\Omega \subseteq \mathbb{C}$ is an open connected set, then are equivalent:

a) $\mathbb{C}_{\infty} - \Omega$ is a connected set.

b) Each piecewise smooth cycle is $\Omega$ - homologue to $0$ .

c) If $\Gamma$ is a piecewise smooth cycle in the open set $\Omega \subset \mathbb{C}$ , then $\forall f \in \mathcal{H}(\Omega)$ and every $z \in \Omega \text{ - Image(}\Gamma)$ : $\displaystyle \text{ Ind (}\Gamma , z) \cdot f(z) = \frac{1}{2i \pi} \oint_{\Gamma} \frac{f(w)}{w - z} \, dw$ .

d) For each piecewise smooth cycle $\Gamma$ in $\Omega$ and each $f \in \mathcal{H}(\Omega)$ , $\oint_{\Gamma} f(z) \, dz = 0$

e) $\Omega$ is holomorphically connected,i.e, $\forall f \in \mathcal{H}(\Omega) \space \exists F \in \mathcal{H}(\Omega$ ) such that $F' = f$

f) $\forall f \in \mathcal{H}(\Omega)$ with $0 \notin f(\Omega) \space \exists g \in \mathcal{H}(\Omega)$ such that $e^{g} = f$

h) $\forall u$ harmonic and analytical function in $\Omega$ there exists a harmonic and analytical function $v$ in $\Omega$ such that $u + iv \in \mathcal{H}(\Omega)$

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – I have fixed the section c) of the previous comment. It's horrible not to be able to preview the posts... Good, I just only need the definitions of singularities and residue theorem... Before giving them, I want to write what I used: $\frac{e^{i \pi x}}{4x^2 - 1} = \frac{e^{i \pi x}}{(2x - 1) \cdot (2x + 1)} = \frac{1}{2} \left(\frac{e^{i \pi x}}{2x - 1} - \frac{e^{i \pi x}}{2x + 1}\right) =$ $= \frac{1}{4} \left(\frac{e^{i \pi x}}{x - 1/2} - \frac{e^{i \pi x}}{x + 1/2}\right)$ and $C_r (t) = re^{it}, t \in [0, 2 \pi]$ , $\Omega = \mathbb{C} - \{1/2, - 1/2\}$ and limits...

Theorem (Residue theorem)

Let $f \in \mathcal{H}(\Omega)$ , $S \subseteq \delta \Omega$ (boundary of $\Omega$ ) a finite set of isolated singularities of $f$ , and $\Omega_{S} = \Omega \cup S$ . If $\Gamma$ is a piecewise smooth cycle in $\Omega$ and $\Omega_{S}$ - homologue to $0$ then $\displaystyle \frac{1}{2i \pi} \oint_{\Gamma} f(z) \, dz = \sum_{a \in S} \text{ Ind(}\Gamma , a) \cdot \text{ Res (f, a)}$

Good, you can attack me now, doubts, etc.. Ah,I need the definition of singularities... Today, I have finished here... Tomorrow,I'll be back

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – I have a last question, why $\int_{- \infty}^{\infty} \frac{e^{i\pi x}}{4x^2 -1} \, dx = \frac{-\pi}{2}.$ ? wouldn't it be $-\pi$ ?(due to residue theorem)... I have added a picture and a line to my proof... the cycle $C_r(t) = re^{it}, \space t \in [0, 2\pi]$ is the union of two cycles... (See the picture) I hope everything is clear now... Please, does there exist one more doubt? Please, be sincere...

Guillermo Templado - 4 years, 6 months ago

@Refaat M. Sayed , is this your intended solution? If it's not, can you share your approach?

Pi Han Goh - 4 years, 6 months ago

I don't understand you very good... This is my solution, I haven't copied from internet

Guillermo Templado - 4 years, 6 months ago

I asked the author of this question whether his technique to solve this question is similar to yours.

Pi Han Goh - 4 years, 6 months ago

@Pi Han Goh – Ah sorry, then @Refaat M. Sayed , please could you delight us with your solution?and if this problem is a question for the rest of people, could you say us where we can find it, please? This is not an order or obligation for you, it's a humble request...

Guillermo Templado - 4 years, 6 months ago

@Pi Han Goh – First of all, I would like to say (because I do not feel embarrassed and I do not care what others think of me) that I have used internet many times to solve problems, sometimes to get points, other times to see the solution and try to learn something ... Well, it does not matter why I did it... @Pi Han Goh , you made this problem, didn't you? If you were right, could you share your solution with us,please? I repeat this is not a order or an obligation... You can do it if you want... If you don't want to do it, don't do it,,, I know that more than 50%( I'm sure) of people than answered this question rightly,they used Wolfram alpha... If it is a lie, they can show it, giving a solution, or everybody made the same solution than me?

Guillermo Templado - 4 years, 6 months ago

@Guillermo Templado – No, I didn't make this problem.

"4" is a very common answer, so I'm pretty certain most people guess the answer correctly.

Pi Han Goh - 4 years, 6 months ago

@Pi Han Goh – maybe you make reason, I don't know... If I had to bet I would bet on Wolfram alpha... 93% of people getting the correct answer on a problem with these characteristics seems to me too much ... althought everything is possible... maybe you make reason.

Guillermo Templado - 4 years, 6 months ago

A calculus problem by Refaat M. Sayed

The answer is 4.

1 solution

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