A very interesting integral solved using my favorite tricks

Рет қаралды 4,675

6 ай бұрын

Here's a fascinating integral with exponential and trig functions solved using Feynman's trick.
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Пікірлер: 35

@nightmareintegral5593 6 ай бұрын

Yeah he changed sign 07:51 Therefore it is π/4 - ln(2)/2 Anyway AMAZING integral! I still love your channel 🖤

@Mr_Mundee 6 ай бұрын

the answer should be pi/4 - ln(sqrt2)

@TheHellBoy05 6 ай бұрын

I got the same but I converted the integral to a double integral

@julianwang7987 6 ай бұрын

Alternatively, Taylor expand cos(x) and the integral can be expressed as the sequence 1/1-1/2-1/3+1/4+1/5-1/6... Utilizing the equality -ln(1-x)=x/1+x^2/2+x^3/3+... and setting x=i, the previous sequence can be shown to be exactly the imaginary part of -ln(1-i)*(1+i), which easily simplifies to pi/4-ln(2)/2.

@Spiderp-p1l 6 ай бұрын

Love it every time you post these integrals, they're the best to watch right before bed:)

@giuseppemalaguti435 6 ай бұрын

Scrivo (1-cosx)/x^2 in forma di serie...I=1-1/2-1/3+1/4+1/5-1/6-1/7+1/8+1/9..=(1-1/3-1/5-1/7+1/9...)-1/2(1-1/2+1/3-1/4..)=arctg1-1/2(ln2)=π/4-ln√2.

@MrWael1970 4 ай бұрын

Thanks for your effort.

@Bedoroski 3 ай бұрын

Can you discuss more about taking the real part outside the Integral? Never learnt this before

@陳柏勳-h8f 6 ай бұрын

also can use Laplace transform f(t)=(1-cos(t))/t^2 F(s)=L{f(t)}=int_s^(inf) [int_s^(inf) L{1-cos(t)}] = int_s^(inf) [int_s^(inf) 1/s-s/(1+s^2)] = int_s^(inf) [ln(s)-1/2*ln(1+s^2)]_s^(inf) = int_s^(inf) 1/2*ln(1+s^2)-ln(s) =[1/2*s*ln(1+s^2)-s*ln(s)+arctan(s)] _s^(inf) =pi/2+s*ln(s)-1/2*s*ln(1+s^2)-arctan(s) F(s)=L{f(t)}=int_0^(inf) exp(-st)*(1-cos(t))/t^2 let s=1 F(1)= int_0^(inf) exp(-t)*(1-cos(t))/t^2 = pi/2+ln(1)-1/2*ln(2)-arctan(1) =pi/4-1/2*ln(2)

@kewalmer7225 6 ай бұрын

I used e^-ax instead of cosax and got the answer as pi/4-1/2(ln2)

@theblainefarm3310 6 ай бұрын

Nice. I am going to feature this integral on my other channel, which focuses almost exclusively on this technique. I will wait a few days so as not to step on your toes though. Love your videos!

@michaelguenther7105 6 ай бұрын

For convergence you checked as x->Infinity, but you should also check as x->0. It works out, but you should show that.

@_ccinfinity 6 ай бұрын

Bruh it's not a real analysis class.

@ambiguousheadline8263 6 ай бұрын

7:53 why is that a plus

@aravindakannank.s. 6 ай бұрын

yeah minor mistake 😅

@abdulazizhamid1607 6 ай бұрын

8:00 should it not be -log(√2), I'm confused

@shivanshnigam4015 6 ай бұрын

Yes definitely

@Bedoroski 3 ай бұрын

Great video

@evanlink2414 6 ай бұрын

I have a suggestion. The triple integral from negative infinity to positive infinity of (2yzsin(y) - yzsin(2x + y) + yzsin(2x - y)) / (4(x^2)(y^2)(e^(z + e^(-z))) + 4(x^2)(e^(z + e^(-z)))) dxdydz. It converges to (gamma*pi^2)/e

@ericthegreat7805 6 ай бұрын

In the arctan(1) = pi/4, note that cos(pi/4) = 1/sqrt(2) and the sqrt(2) appears again. Of course you have the usual suspect log 2. The cosine comes from 1 - cosx. I want to know if anyone else has looked at integrals in terms of information like this in relation to log 2.