A ridiculously awesome integral with an epic result

[@rajendramisir3530]

A reliable heterogeneous mixture that leads to a succulent solution of this esoteric integral. Clever choice of tools, techniques and clear explanation. Thanks for sharing. Recognizing and applying identities is a very helpful technique to solve problems in Mathematics.

[@maths_505]

Spoken like a poet

[@fartoxedm5638]

I just discovered your channel yesterday, but I'm already in love! You remind me of my teacher, who always left out some details, but he always mentioned them to make sure that we also understand all the basic things.

[@michaelbaum6796]

You are really the best teacher I have ever had. Your videos are so clear. I live them.

[@violintegral]

AMAZING integral. We get the trifecta of famous mathematical constants: e, pi and phi all in relation with one another through this integral. I found another great solution using the series expansion of cosine, the Gamma and Beta function, and the geometric series but I don't want to step on your toes by sharing too many of my own solutions. It's your channel, after all.

[@maths_505]

I found that solution too but this technique is way more awesome. After all, the only way to make integrals even more entertaining is to throw complex numbers, integral transforms and special functions into the mix😂

[@rajendramisir3530]

13:32 A reliable heterogeneous mixture that leads to a succulent solution of this esoteric integral. Clever choice of tools, techniques and explanation. Thanks for sharing. Recognizing and applying identities is a very useful technique to help construct solutions to problems in Mathematics.

[@tueur2squall973]

May I ask you how did you achieve the same results by using at first the geometric series of cosine ? I'm trying my best but I'm stuck. Here's my work for the moment : Step 1: transform the integral from -infty to+infty into 2*integral from 0 to +infinity because our integrand is even. Step 2: Write cos(2x²) as its geometric series Step 3: slip the exponential term inside the sum and switch up the sum and the integral Step 4 : small change of variable to get the gamma function. We let t=x² Step 5 : We have finally the sum over k of [(-4)^k]Gamma(2k+1/2)/(2k)! Then what do we do next ?

[@violintegral]

@@tueur2squall973 multiply by Gamma(1/2)/Gamma(1/2). Then the summand is 1/sqrt(pi)*(-4)^n*Gamma((4n+1)/2)*Gamma(1/2)/Gamma(2n+1). The Gamma function expression is of the form Gamma(x)*Gamma(y)/Gamma(x+y) = B(x, y), so you can use an integral representation of the Beta function to transform the sum into an integral. By the way, the resulting geometric series has common ratio (-4x^2), and since the integral bounds are from 0 to 1, it does not converge in the usual sense. Now I don't know anything about complex analysis, but I believe it's valid to analytically continue the geometric series here so that we can write the sum as 1/(4x^2 + 1). And, indeed, this approach seems to be valid since the resulting integral gives the same value found in this video. However, evaluating the resulting integral is not an easy task either, so good luck!

[@TheArtOfBeingANerd]

I love how the radical radical 5 at around 9:45 looks like it's just morphing

[@omar990anbar]

It was a good plot twist at the very end. Golden Ratio: have you forgotten me boys?

[@qetzing]

I am simply amazed. I don't have the words to express how elegant the solution and even more so your presentation is. Truly perfect

[@maths_505]

Thanks mate

[@michaelbaum6796]

Fascinating solution👍It‘s so much fun watching your perfect videos. Go ahead!😀

[@maths_505]

Thanks bro

[@maalikserebryakov]

Its always a spiritual experience seeing your solutions. I feel my soul being thrown out of my body and through the cosmos. I see everything for a few seconds.

[@maths_505]

Bro🥺

[@renesperb]

It seems worthwhile to show that for arbtrary complex a with positive realpart one has ∫ exp[-a*x^2] dx from 0 to inf = 1/2*√(π/a).This includes a number of special cases , like the one discussed in this video.

[@tsa_gamer007]

does anyone knows that that in I(a)= the integral from 0 to infinity of e^(-a(x^2)) dx put ax^2=t => I(a)=1/(2sqrt of a) * gamma function of (1/2) so for a=1-2i we get our integral= Re(sqrt[pi/1-2i]) making the denominator real we get sqrt(pi/5) Re(sqrt(1+2i)) =sqrt(pi/5) sqrt golden ratio

[@newwaveinfantry8362]

Love it! ♥

[@AhmadAli-fx1hk]

The end result was pretty amazing

[@erggish]

I have no idea how you got the results from the Laplace (and inverse) operators... But the end result of pi, phi and five was great indeed :D

[@manstuckinabox3679]

Involving my favorite Identity?? SWEET!

[@chenwong1036]

Where do these integral come from and how are these used in real life applications?

[@manstuckinabox3679]

Reviewing this, 2:36, just as I was about to ask LOL.

[@fonaimartin98]

Subbing u = x sqrt(1 - 2j) into the integral (of which's real part we are interested in) also works, doesn't it? (Assuming we take the value of the Gaussian integral as known)

[@maths_505]

I wanted the proof of the gaussian integral for a complex argument to be part of the video as kind of an added bonus as I found the evaluation quite beautiful

[@carlosdavid7430]

I got an 8 on my spanish exam,can you give me some integrals i can solve to cheer me up?

[@maths_505]

Let me think of some to post as HW Sorry about the Spanish exam mate....you'll get em next time

[@rajibdebnath9896]

Hello sir, can you suggest me good mathematical physics book.

[@maths_505]

Oh there are lots of great books... For introductory concepts I recommended Stone and Golbert or Sadri Hassani's mathematical physics. For specific topics you can check out Sean Carroll's lecture notes in general relativity and there are plenty of resources for mathematics required for quantum mechanics: my favourite QM book by Lifshitz and Landau is full of wonderful mathematical explanations

[@rajibdebnath9896]

@@maths_505 thank you sir.

[@giuseppemalaguti435]

Quando sei arrivato a 2 30 basta utilizzare l'integrale della gaussiana...ma immagino te ne sarai accorto

[@maths_505]

Sì, certo, ma volevo includere la valutazione della gaussiana per un argomento complesso come parte dello sviluppo della soluzione per l'integrale perché era estremamente interessante

[@holyshit922]

Laplace transform yes it is good idea but Euler's formula can be avoided , nevertheless nice solution