You may critisize your hand writing but your speech is excellent. Even for me as non native speaker I understand almost every word in your videos. Sometimes for the jokes I have to switch on subtitles or if you use German math terms. But anyway Leibnitz is pronounced perfectly by you, which is not natural with this diphtong😊

Mathematica gets it (after a while, producing the equivalent form (-1 + Sqrt[1/2 + 1/Sqrt[2]]) Pi. But is great fun by hand, with residues and all. (partial fractions, etc...)

@maths_505

3 ай бұрын

Yes indeed

That second sub was crazy!!

An indefinite integral can be found, which you proved by finding a rational function. Then you use limits to find the value of the indefinite integral at infinity. It’s time-consuming, but possible.

Awesome video ❤

I = int sqrt(sin*cos + sin^2(x))/cos x => int sqrt(tan x + tan^2 x) dx => int sqrt(u+u^2)/(u^2+1) => (u+u^2) * arctan u - int arctan x + 2x arctan x dx

Very smart solution, but i expect to solve the integral in real domain. So, thank you for your featured effort.

@maths_505

3 ай бұрын

I will always take the complex route whenever I can 😂

Yes you can use real analysis to solve that rational function integral but it involves some really hairy trig manipulations and/or some truly gnarly partial fractions.

For the I1, it's an even function, can't you say it's half the integral from -inf to inf and then substitute t+1 to get an integral of the form of beta functions? I haven't tried it myself but I believe so.

@maths_505

3 ай бұрын

Nope that won't work.

18:44 we use " 1er lemme de jordan " to prove directly that the integrale here is equal 0 juste if the arc of cercle does not encounter any pole if R tends towards infinity... is that true ? Cause we study that here in morocco .

@maths_505

3 ай бұрын

Nah the integrand is suitable for the ML estimation lemma rather than the Jordan's lemma for which we need a complex exponential term as part of the integrand.

Would you please suggest me any lecture or book where i can learn 'complex analysis' it will we very helpful for me 🙏🙏

@maths_505

3 ай бұрын

My complex analysis lectures are linked in the description

@Flash_098

3 ай бұрын

Ok sir , Thank you ❤

at 1:08 should that not be dx = du/(1+u^2) ? did dx and du switch or is this just a typo?

@maths_505

3 ай бұрын

Yes it's a typo

Hey! Would you want to solve some interestingly convoluted PDEs within the week ? Would love it. Love you brother

7:37 divide the numerator and denominator by t²

Is this your solution? If so, how do you know you’re making any progress on a problem this hard? and how do you know how to start?

@josephlorizzo8997

3 ай бұрын

yeah, that'd be really useful info, i Also Always wandered the same thing

@maths_505

3 ай бұрын

Yup it's my solution development. Progress? Well if the math keeps working and I get things I can solve then that's definitely progress. Where to start is harder to explain since alot of it is gut feeling at this point. The more integrals you solve the better is my advice.

@josephlorizzo8997

3 ай бұрын

@@maths_505 thank you buddy!!! love your vids so much, the integrals you solve are some of the funniest, thank you very much

For I sub1 you could have divided the numerator and denominator by 1/t^2 and you will get a nice arctan structure. good content keep up the work

@maths_505

3 ай бұрын

Tried but it didn't work.

@hamdansaad3406

3 ай бұрын

@@maths_505 ahh you are right I made a mistake ty

Could you please solve this integral using beta and gamma function ??

@maths_505

3 ай бұрын

Bro I can't

hey if we calculate the value of this integral numerically (i used python) and it comes out to be 0.310025285896493 can anyone prove to me this value for the end result??

@amritlohia8240

3 ай бұрын

If you type the final result from this video into a calculator, you'll indeed see that it matches your numerical value.

Ooh a complex snac

they asked this in the 2021/22 uk integration bee and I got destroyed

hey it's me again and i got and other interage that i think have a nice result, it's the integrale for 0 to inf of (e^-(x^2)-e^-x)/x i have try using feyman trick but nah i can't, and i don't want to play with lim of Ei(x), so if u i can give u a video idea i'lll be happy to watch it haha (and btw i watch the other one with xln(1+cosx) i learn a great trick for ln(cosx)

@SussySusan-lf6fk

24 күн бұрын

I've a very elegant solution for you. Read it step by step. Let, I(a)= int 0 to inf (e^(-ax^2) - e^(-ax)) /x dx I'(a) = 1/2a int 0 to inf. - 2ax e^(-ax^2) dx + int 0 to inf e^-(ax) dx I'(a) = 1/(2a) I(a)= 1/2 ln a +c Now we put a=i =sqrt(-1) I(i)= int 0 to inf. (e^(-ix^2) - e^(-ix))/x dx = int 0 to inf, { cos(x^2) - isin(x^2) - cosx + isinx} /x dx = int 0 to inf, {cos(x^2) - cosx} / x dx + i int 0 to inf, (sinx - sinx^2) /x dx We know sinx/x dx from 0 to infinity = pi/2, and sinx^2 / x dx from 0 to infinity = pi/4, we can just substitute x^2=u to get that =int 0 to inf, {cos(x^2) - cosx} / x dx + i pi/4 Let g(t) =int 0 to inf, {cos(tx^2) - cos(tx) } / x dx Laplace of g(t) =L(g(t))= int 0 to infinity e^(-st) [int 0 to inf, {cos(tx^2) - cos(tx) }/x dx ] dt =int 0 to inf 1/x ( int 0 to inf, e^(-st) {cos(tx^2) - cos(tx)} dt ) dx = int 0 to infinity 1/x { s/(s^2 +x^4) - s/(s^2 +x^2)} = - int 0 to infinity - 1/(4s) 4/x^5 s^2 /((s^2/x^4) - 1) dx + int 0 to infinity - 1/(2s) 2/x^3 s^2 /((s^2/x^2) - 1) dx = [1/2s ln{(s^2 + x^2) /sqrt(s^2 + x^4)} ] put limits first inf then 0 = - 1/2 lns /s (g(t)) = - 1/2 L^-1 (lns/s) = y/2 + lnt /2, y=eular masheroni constant g(1)= y/2 = int 0 to inf, {cos(x^2) - cosx} / x dx So, I(i) = y/2 + i pi/4 We already had, I(a) =1/2 ln a + c I(i ) =y/2 + i pi/4 =1/2 ln i + c y/2 + i pi/4 = c + i pi/4 {lni = i pi/2, as principle value} c=y/2 I(a) = 1/2 ln a + y/2 I(1)= y/2, you can verify it for any a

@sachacloot1925

20 күн бұрын

@@SussySusan-lf6fk WOW that's amazing to think about calculate I(i) to find "c" am very impressed thank u !!

@Mathematician6124

20 күн бұрын

​@@SussySusan-lf6fk😊 Very nice. Susan!

If it's MUCH harder than it looks then to an idiot like me that would imply it's impossible.