If we introduce I(b) = int_{-infty}^0 e^y/y * sin(b*y) dy, we get dI(b)/db = int_{-infty}^0 e^y * cos(b*y) dy, which evaluates straightforwardly to 1/(1+b^2). (basically the Laplace transform of cos(b*t) evaluated at s=1 after substituting t=-y) Therefore I(b) = arctan(b) + C. And since I(0) = arctan(0) = 0, we have C = 0, giving us I(1) = arctan(1) = pi/4.
No need Laplace transform, y->exp(y) cos(by) has a simple antiderivative computed by two integration by parts. d(sin(by)/b)= cos(by) and then d(-cos(by)/b^2)=sin(by)/b
Without watching the video: Let I(a) = int from 0 to 1 of sin(aln(x))/ln(x) With Feynman's trick, and expressing cosine as a complex exponential, we easily obtain I'(a)=1/(1+a²). With I(0)=0, which is easy to note from I(a)'s definition, we get I(a) = arctan(a). The target case with a=1 gives I(1) = π/4 as the answer
I did it by writing sin(alnx)/lnx with feynman's trick the derivative becomes cos(alnx)=Re(e^(ialnx))=Re(x^(ia)). Then integrate, to get Re(1/(1+ia))=Re((1-ia)/(a^2+1)=1/(a^2+1). Integrate to give arctana+C. Setting a=0 gives the integral=0 because sin0=0. Therefore, we get arctana and setting a=1 we get arctan1=pi/4.
Can integrate also f(z)=exp(-z)/z on a contour with 1) arc of radius epsilon centred at 0 2) arc of radius R centred at 0 3) line from R*exp(i pi/4) to epsilon*exp(i pi/4) 4) line from epsilon*exp(-i pi/4) to R*exp(-i pi/4)
We could calculate in C with sin(z) = (e^iz - e^(-iz) / 2i and get 1/2i*[Ei((1+i)lnx) - Ei((1-i)lnx)] and look at the difference of the power series of the Ei-functions, leaving us with: (-i/2*ln((1+i)/(1-i)) which is arctan(1) = pi/4
Clever! I did this integral less cleverly but I still got the answer, which doesn't usually happen with these integral videos. I also did the substitution, which gives the integral of sin(y)e^(y)/y dy from -∞ to 0, but then I got a bit stuck (because I never think of differentiating under the integral sign) so I just... turned sin(y)/y into an infinite series, the sum from n = 0 to ∞ of (-1)^n · y^(2n)/(2n + 1)!. Then it's just integration by parts. If f(n) = ∫y^(n)·e^y dy from -∞ to 0, you can easily see that f(0) = 1, and by applying integration by parts, you get that f(n) = -n·f(n - 1), so f(n) = (-1)^n · n!. This eventually gets you to the integral as 1 - 1/3 + 1/5 - 1/7 + ..., which is a series that converges to π/4 *famously* slowly. I feel like I got lucky that I remembered that series. Not sure I would have been able to get π/4 from that otherwise.
@M.Z.M.N. sin(y)=(exp(iy)-exp(-iy))/(2i). and then you apply the same differentiation trick as in the video, but you only need to integrate a simple exponential function
For integral lovers this one’s a little formulaic, forced u sub u=ln(x) makes the integral become a quantity that comes up when you apply Feynmans trick to sin(x)/x, so you take what you know there and just rinse and repeat.
It's simpler if you parametrize I(z)=integral sin(y)exp(-zy)/y and then express I'(z) which you can find easily by expressing sin(y) in exponential terms so that I'(z)=1/(1+z^2). Then you find the integration constant by making z->infinity and it's already done !
This integral was definitely an interesting challenge. I love how different approaches can lead to the same result. I’ve been using SolutionInn's AI tools to help me with tricky integrals like this, and it’s been a huge help to get through the steps when I get stuck or need a fresh perspective. This series approach is a clever twist, and it’s great to see how it all links back to something as iconic as π/4!
Why do you always skip over the details of if/how to change the order of integration. That's an interesting and important step. You have at least said "by Fubini's Theorem" before, but it think it would be important and interesting to go through the steps.
@ sure! First you do the substitution that he did and convert the integral to that of -(sin(x)/x)e^(-x) from 0 to inf this is the laplace transform of sinx/x and from the properties of the laplace transform we know that L(f(t)/t) = L(f(t)) integrated from s to inf thus the result is the integral of 1/(x^2+1) from s to inf with s=1
0:04 Bless you
0:07 Actual start
4 seconds of hard work, not gonna throw that away, we have come too far.
Help, please. I couldn't find a good place to stop and now I'm lost.
I feel so LOST
Law of inertia bro. Call Newton to apply force
If we introduce I(b) = int_{-infty}^0 e^y/y * sin(b*y) dy, we get dI(b)/db = int_{-infty}^0 e^y * cos(b*y) dy, which evaluates straightforwardly to 1/(1+b^2).
(basically the Laplace transform of cos(b*t) evaluated at s=1 after substituting t=-y)
Therefore I(b) = arctan(b) + C. And since I(0) = arctan(0) = 0, we have C = 0, giving us I(1) = arctan(1) = pi/4.
No need Laplace transform, y->exp(y) cos(by) has a simple antiderivative computed by two integration by parts. d(sin(by)/b)= cos(by) and then d(-cos(by)/b^2)=sin(by)/b
4:15 After getting the dz integral we can use exponential formula for sin and immideately get the last integral using the fact \int_0^inf e^(az)=1/a
Without watching the video:
Let I(a) = int from 0 to 1 of sin(aln(x))/ln(x)
With Feynman's trick, and expressing cosine as a complex exponential, we easily obtain I'(a)=1/(1+a²).
With I(0)=0, which is easy to note from I(a)'s definition, we get I(a) = arctan(a). The target case with a=1 gives I(1) = π/4 as the answer
I did it by writing sin(alnx)/lnx with feynman's trick the derivative becomes cos(alnx)=Re(e^(ialnx))=Re(x^(ia)). Then integrate, to get Re(1/(1+ia))=Re((1-ia)/(a^2+1)=1/(a^2+1). Integrate to give arctana+C. Setting a=0 gives the integral=0 because sin0=0. Therefore, we get arctana and setting a=1 we get arctan1=pi/4.
At 5:20 I'd have replaced sin(y) with imaginary part of exp(iy) and gone on from there.
Can integrate also f(z)=exp(-z)/z on a contour with
1) arc of radius epsilon centred at 0
2) arc of radius R centred at 0
3) line from R*exp(i pi/4) to epsilon*exp(i pi/4)
4) line from epsilon*exp(-i pi/4) to R*exp(-i pi/4)
Feynman’s trick or something
Feynman's trick is pretty much the same thing as what Michael did.
We could calculate in C with sin(z) = (e^iz - e^(-iz) / 2i and get 1/2i*[Ei((1+i)lnx) - Ei((1-i)lnx)] and look at the difference of the power series of the Ei-functions, leaving us with:
(-i/2*ln((1+i)/(1-i)) which is arctan(1) = pi/4
Feymanns trick or contour integration seems good
Clever! I did this integral less cleverly but I still got the answer, which doesn't usually happen with these integral videos. I also did the substitution, which gives the integral of sin(y)e^(y)/y dy from -∞ to 0, but then I got a bit stuck (because I never think of differentiating under the integral sign) so I just... turned sin(y)/y into an infinite series, the sum from n = 0 to ∞ of (-1)^n · y^(2n)/(2n + 1)!. Then it's just integration by parts. If f(n) = ∫y^(n)·e^y dy from -∞ to 0, you can easily see that f(0) = 1, and by applying integration by parts, you get that f(n) = -n·f(n - 1), so f(n) = (-1)^n · n!. This eventually gets you to the integral as 1 - 1/3 + 1/5 - 1/7 + ..., which is a series that converges to π/4 *famously* slowly. I feel like I got lucky that I remembered that series. Not sure I would have been able to get π/4 from that otherwise.
i think writing sin(y) in exponential form gives the same result a bit quicker.
How??
@M.Z.M.N. sin(y)=(exp(iy)-exp(-iy))/(2i). and then you apply the same differentiation trick as in the video, but you only need to integrate a simple exponential function
Oof, that splitting up into 2 terms trick came out of nowhere. Idk how I ever would have thought to do that.
It seems to have been crafted from wanting to cancel 1/z and z, after noticing they both appear in the integration by parts
For integral lovers this one’s a little formulaic, forced u sub u=ln(x) makes the integral become a quantity that comes up when you apply Feynmans trick to sin(x)/x, so you take what you know there and just rinse and repeat.
It's simpler if you parametrize I(z)=integral sin(y)exp(-zy)/y and then express I'(z) which you can find easily by expressing sin(y) in exponential terms so that I'(z)=1/(1+z^2). Then you find the integration constant by making z->infinity and it's already done !
But you need to use complex numbers.
@@richardheiville937 Yes but often using complex numbers is less complex 😉
What did you do in the third step when you introduced "z" into the integral?
0:04 You should do more backflips again to relax your muscles. 😮
😏
bless ya
t = -ln(x)
L(f(t)/t) where f(t) = sin(t)
then plug in s = 1
Another way to do it:
sin(log(x)) = (x^i - x^(-i))/2i
I(p) = int from 0 to 1 (x^(p*i) - x^(-i))/log(x) dx
And there you go.
Why this? Can you elaborate?
@@solcarzemog5232 sin(x) = (e^(ix) - e^(-ix))/(2i). This is a bit loose but correct application of complex analysis.
@@rotoboravtov4354 yea, my bad 😅
So here's the thing you still need feymmans technique, or you need to turn that into a tripe integral to solve it
It makes no difference
Need a good place to start :)
This integral was definitely an interesting challenge. I love how different approaches can lead to the same result. I’ve been using SolutionInn's AI tools to help me with tricky integrals like this, and it’s been a huge help to get through the steps when I get stuck or need a fresh perspective. This series approach is a clever twist, and it’s great to see how it all links back to something as iconic as π/4!
I don't understand, it's a beautiful feeling, I wanna understand so much!
Breathing chalk dust too many years!!!
Looks like sin(u)e^u/u du to me. That u denominator looks rough though. Maybe there is a definite-integrals-only trick too. 🤔
B-b-but where's the good place to stop?
Why didn’t he just restart filming 😂
And that's a good place to stop
Why do you always skip over the details of if/how to change the order of integration. That's an interesting and important step. You have at least said "by Fubini's Theorem" before, but it think it would be important and interesting to go through the steps.
not an easy one; many non obvious stuff to imagine here...
Lol you way overcomplicated it. Could have easily solved it using the properties of the laplace transform.
Can you teach me how to do that? I only have learned a little about it when we use it to solve differential equations
@ sure! First you do the substitution that he did and convert the integral to that of -(sin(x)/x)e^(-x) from 0 to inf this is the laplace transform of sinx/x and from the properties of the laplace transform we know that L(f(t)/t) = L(f(t)) integrated from s to inf thus the result is the integral of 1/(x^2+1) from s to inf with s=1
Ok , I kinda get it ! Thanks!
@EliavShalev -- Write sin(x)/x.
Okay. That was like a f*cking magic trick :D
Get rid of that major curse word. It's ignorant and needless.
@@robertveith6383 no