Yeah I think when you got to step that has e^(e^(2*i*x)), I substituted the definition of sine function in terms of complex exponentials. From there you can make a sub that z = e^(i*x), and then use residue theory on the resulting function. We can treat this result as if we integrated around a closed contour, maybe a circle with r > 1, this should give you 2*pi*e*i, then just find a quarter of this result and take the imaginary part? I think there might need to be more arguments for those last steps, but I think it’s a simple as your argument near the end, you get double the result if you integrate from [-pi/2, pi/2], but because the integrand is also periodic, this interval is the same thing as saying [0, pi], so from [0, 2pi], is just the other half of that result. Cool problem!
Dear sir, I am a highschooler and I have much trouble with understanding When you can and can not switch order of operations in integrals & summations. I have checked wikipedia but the kind of math notation is a bit above my understandings, could you help me out? Or do you know a good site to learn this? Love ur vids, I try to understand everything but this is just getting in the way now :)))
It all depends on convergence. There are lots of ways to check convergence like I did in this video. For integrals its much easier cuz you can always look at a graph in wolfram alpha. For double sums it's a bit more difficult as you have to check for absolute convergence of the iterated sum. Check out math stackexchange, you'll find lots of comprehensive explanations.
Hey math505, (unrelated question) which books from springer do u recommend a student who is going to university.... the first topic is infinite sequence and series....
Also, I've completed the following chapters: functions ,limits ,continuity and differentiability, differentiation and its application, indefinite and definite integration and area under curve....which topic should i start next?
When I saw Im [e^(e^(2ix)) e^(ix)] / sin(x), I thought of rewriting it as Im [2 i e^(e^(2ix)) e^(ix)] / [e^(ix) - e^(-ix)] dx = 2 Im [i e^(e^(2ix))] / [1 - e^(-2ix)] dx As is mentioned in the video, because the original function is even * odd(odd) / odd, the resulting function is an even function. I then thought to make a substitution of z = e^(2ix), making dz = 2i z dx, or dx = dz / (2i z) This makes this 2 Im[i (e^z / (1-1/z)) / 2iz] Applying that this is an even function, we get Im(e^z / (1-z) dz)/2. This gives us that we can use residue theory to get the at the residue of z = 1 of simply e/2, and therefore, the integral is simply pi*e.
That was awesome artistry as usual. (BTW I was wondering, if you had time to tell me if this is this a legitimate use of the gamma function: ua-cam.com/video/7_iWDyMPoJs/v-deo.html
Euler's wonderful *buuuurp* wonderful formula, Morty
Very nice! The one who came up with this integral is so creative!💯
Loved your previous result with the old Euler-Mascheroni and what not. This looks like another spicy boi!
Why is it so hard for me to imagine Rick doing actual maths
omg beautiful solution
Very Smart Solution. Good job.
Yeah I think when you got to step that has e^(e^(2*i*x)), I substituted the definition of sine function in terms of complex exponentials. From there you can make a sub that z = e^(i*x), and then use residue theory on the resulting function. We can treat this result as if we integrated around a closed contour, maybe a circle with r > 1, this should give you 2*pi*e*i, then just find a quarter of this result and take the imaginary part? I think there might need to be more arguments for those last steps, but I think it’s a simple as your argument near the end, you get double the result if you integrate from [-pi/2, pi/2], but because the integrand is also periodic, this interval is the same thing as saying [0, pi], so from [0, 2pi], is just the other half of that result.
Cool problem!
Oh yeah I figured it out, the function is also periodic so you use that to make the simplification.
Solving along the vid and finally getting the answer pulls out a *WOAHHHHHHH* that'll put 7 year olds to shame lol. Awesome vid thanks !
Dear sir, I am a highschooler and I have much trouble with understanding When you can and can not switch order of operations in integrals & summations. I have checked wikipedia but the kind of math notation is a bit above my understandings, could you help me out? Or do you know a good site to learn this? Love ur vids, I try to understand everything but this is just getting in the way now :)))
It all depends on convergence. There are lots of ways to check convergence like I did in this video. For integrals its much easier cuz you can always look at a graph in wolfram alpha. For double sums it's a bit more difficult as you have to check for absolute convergence of the iterated sum. Check out math stackexchange, you'll find lots of comprehensive explanations.
@@maths_505thx i now understand it well, also the example in this video is clear now !!
Hey math505, (unrelated question) which books from springer do u recommend a student who is going to university.... the first topic is infinite sequence and series....
Also, I've completed the following chapters: functions ,limits ,continuity and differentiability, differentiation and its application, indefinite and definite integration and area under curve....which topic should i start next?
Real analysis by Rudin (also know as baby rudin) is an absolute classic
@@maths_505 Thankyou ❤
What's wrong with extending the imaginary part argument to csc(x)?
*I LIKE WHAT YOU GOT!*
now try to find an integral which result is e times pi times gamma
nice🙂
When I saw Im [e^(e^(2ix)) e^(ix)] / sin(x), I thought of rewriting it as Im [2 i e^(e^(2ix)) e^(ix)] / [e^(ix) - e^(-ix)] dx = 2 Im [i e^(e^(2ix))] / [1 - e^(-2ix)] dx
As is mentioned in the video, because the original function is even * odd(odd) / odd, the resulting function is an even function.
I then thought to make a substitution of z = e^(2ix), making dz = 2i z dx, or dx = dz / (2i z)
This makes this 2 Im[i (e^z / (1-1/z)) / 2iz]
Applying that this is an even function, we get Im(e^z / (1-z) dz)/2.
This gives us that we can use residue theory to get the at the residue of z = 1 of simply e/2, and therefore, the integral is simply pi*e.
I thought exactly the same thing
Noise
You love Taylor series, don’t you? Haha just messing with you, nice video!
Ofcourse I love em!
They're incredibly useful
Infinite series in general are super handy
So the answer is 9/2 😅
Yes exactly
That was awesome artistry as usual. (BTW I was wondering, if you had time to tell me if this is this a legitimate use of the gamma function: ua-cam.com/video/7_iWDyMPoJs/v-deo.html
It's legit