Excellent video 👏👏👏 Instead of Feynman's you can use a fundamental property of Laplace Transform - multiplication with 1/x^2 corresponds to double integration in S domain🍻

Merry Christmas 🎁🎄 everyone across the globe 🌎🌍

A tough one!

Another Method ---- ln(x) = t , then diffrentiate 1/x = dt then as x = e^t, we have got the eqn t(e^t/{e^t+1}) then can we just use the ILATE rule to integrate taking t as algebraic and rest as exponential (int of lnt using 1.lnt...

Mammoth

Finally more straightforward than splitting the whole thing into two series expressed with the digamma function. Nice, as usual! 😊

Excellent telescoping method. One small note: The individual infinite sums diverge and strickly speaking cannot be used, though the result will be the same if partial sums were used instead.

Very nice!

Perfecto👏👏👏

x->1/x I=int[1,♾️](floor(sqrt(1+x))/x^2)dx u=x+1 I=int[2,♾️](floor(sqrt(u))/(u-1)^2)du t=sqrt(u) du=2tdt I=2•int[sqrt(2),♾️](t•floor(t)/(t^2-1)^2)dt I=2•int[sqrt(2),2](t/(t^2-1)^2)dt+2•sum[k=2,♾️](k•int[k,k+1](t/(t^2-1)^2)dt) int(2t/(t^2-1)^2)dt=1/(1-t^2)+C I=(1/(1-4)-1/(1-2))+sum[k=2,♾️](k/(k^2-1)-1/(k+2)) k/(k^2-1)=1/2•(1/(k-1)+1/(k+1)) I=2/3+1/2•sum[k=2,♾️](1/(k-1)+1/(k+1)-2/(k+2)) I=2/3+1/2•sum[k=1,♾️](1/k+1/(k+2)-2/(k+3)) I=2/3+1/2•(¥(4)-¥(3)+¥(4)-¥(1)) I=7/4 ¥(3)=3/2-ř ¥(4)=1/3+3/2-ř

Nice job. That was a real trig workout!

Awww yeah, luv me sum Trig!

that is why i always recommend everyone to revise trignometric properties of functions before doing integrals.

That was a very interesting problem. Am I correct in assuming that "regular" college calculus I / II / III would not prepare you to solve a problem like this? In other words, are you using secret techniques that college professors don't want us to know about?

ẞ(u,v)=int[0,♾️](x^(u-1)/(x+1)^(u+v))dx x=z^2 dx=2zdz ẞ(u,v)=2•int[0,♾️](z^(2u-1)/(z^2+1)^(u+v))dz I=1/2•ẞ(3/4,5/4)=Ř(3/4)Ř(5/4)/2Ř(2) I=Ř(1/4)Ř(3/4)/8 I=pi/8•csc(pi/4) I=pi/4sqrt(2)

x=sinh^2(t) dx=sinh(2t) sqrt(x+1)=cosh(t) I=int(sinh(2t)e^-(pi•t))dt I=int(e^(2-pi)t-e^(-2-pi)t)dt I=e^(2-pi)t/(2-pi)+e^-(2+pi)t/(2+pi)+C I=e^-(pi•t)•((2+pi)e^2t+(2-pi)e^-2t)/(4-pi^2)+C I=e^-(pi•t)•(4cosh(2t)+2pi•sinh(2t))/(4-pi^2) cosh(2t)=cosh^2(t)+sinh^2(t)=2x+1 sinh(2t)=2sinh(t)cosh(t)=2sqrt(x)sqrt(x+1) e^-t=cosh(t)-sinh(t)=sqrt(x+1)-sqrt(x) I=(sqrt(x+1)-sqrt(x))^pi•(8x+4+4pi•sqrt(x^2+x))/(4-pi^2)+C

I=int[0,pi/2](cbrt(tan(x))/(1+sin(2x)))dx t=tan(x) dx=dt/(1+t^2) sin(2x)=2t/(1+t^2) I=int[0,♾️](cbrt(t)/(1+2t+t^2))dt I=int[0,♾️](t^(1/3)/(1+t)^2)dt ẞ(x,y)=int[0,♾️](p^(x-1)/(1+p)^(x+y))dp I=ẞ(4/3,2/3)=Ř(4/3)Ř(2/3)/Ř(2) Ř(2)=1!=1 Ř(4/3)=1/3•Ř(1/3) I=Ř(1/3)Ř(2/3)/3 Ř(x)Ř(1-x)=pi•csc(pi•x) I=pi/3•csc(pi/3) I=2pi/3sqrt(3)

That one substitution changed whole game

Suppose we want to play with orthogonalization and we define inner product in the form \int_{-1}^{1}p(x)q(x)\cdot \frac{1}{\sqrt{1-x^2}}dx then p(x)q(x) = can be expressed as sum \sum_{k=0}^{m+n}a_{k}x^{k} so we have integral \int_{-1}^{1}\frac{x^{k}}{\sqrt{1-x^2}}dx to calculate Now we can use substitution x = cos(t) to get integral \int_{\pi}^{0}\cos^{k}{t}\cdot\frac{1}{\sqrt{1-cos^{2}{t}}}(-\sin{t})dt =\int_{0}^{\pi}\cos^{k}{t}dt So is we want to orthogonalize basis of polynomials to get Chebyshov polynomials we need to calculate \int_{0}^{\pi}\cos^{k}{t}dt and then we should set a = 0 and b = \pi in your integral And this reduction can be derived by parts using Pythagorean trigonometric identity

An alternative method would be to substitute x=sinh^2 u, although it's a bit messier since you need to do some simplifications with arcsinh. Although, now that I think about it, it really is just equivalent to setting x=(u-1/u)^2/4, which is what was done in the problem.

Fantastic

You could have directly differentiated u and then solve for root(x+1) and root (x) by adding and subtracting the equations, u= root(x+1)-root(x) , and (1/u)= root(x+1)+root(x)

So if you follow this method and integrate (e^x)(sinx), you would get (e^x)(sinx)(1-1+1-1+1-1+...) - (e^x)(cosx)(1-1+1-1+...) Doing it the normal way you would have (1/2)(e^x)(sinx) - (1/2)(e^x)(cosx) This implied the series (1-1+1-1+1-1+...) is 1/2

Just to simplify a little. Let v be the '+' version of u. uv=1 and v-u=2 sqrt (x) so 4x = v^2-2uv+u^2 = 1/u^2 -2 + u^2 From there 4 dx is easy to see -1/u^3+2u times du I don't know if I'd call that a huge game but it does simplify a little

Nice job. It simplified nicely, but getting there was not straightforward.

@EdwardSileo 8 months ago I have to comment here. This is an excellent video: (1) It explains how inverses work - f(f-1(x)) and f-1(f(x)). (2) It explains the symmetry of inverse functions. (3) It explains the domain issues with the Lambert function (with the correction he mentions below) Thanks. My comment: Today, I saw a video on LF where someone states clearly that LF (W) is NOT A FUNCTION! -despite the name. You can try to arrange things, but it is not a function. Because a function cannot have two values for a given x. Many Thanks

My solution t^3 = tan(x) substitution then by parts with u = t , dv = 3t^2/(t^3 + 1)^2dt After integration by parts 1/(1+t^3) = 1/2 *1/(1+t^3)+1/2*1/(1+t^3) then in one of the integrals 1/2*1/(1+t^3) use t = 1/u substitution then add both integrals

if we use 2/3 as x wont the answer be negative?

If e^x = 1 + x^1/1! + x^2/2! + x^3/3! + ..., then wouldn't... i^1/1! + i^2/2! + i^3/3! + ... = e^i - 1? And if e^ix = cos(x) + i*sin(x), then with x=1, wouldn't e^i - 1 = cos(1)-1 + i*sin(1) = -.45 + 0.84i?

This can be mcqs material

Excellent

I wish you used the first method. I find all these formulae overwhelming. I still love it.

Consider the lengths of the intervals over which the integrand is equal to n. n = 1 is a special case where we integrate from 1/3 to 1. Otherwise the length of the interval is 2n+1/((n+2)(n+1)n(n-1)) and the value of the integral is 2n+1/((n+2)(n+1)(n-1)). Using partial fraction decomposition the value of the integral over each interval is -1/(n+2) + 1/2 1/(n+1) + 1/2 1/(n-1). Take the sum from n = 2 to infinity, then re-index to get the sum from n = 4 of -1/n the sum from n = 3 of 1/2 1/n and the sum from n = 1 of 1/2 1/n. Then all the terms with n >= 4 sum to zero and the remaining terms plus 2/3 sum to 7/4.

Very enjoyable! Thank you! Back when I was first learning math I did not care for analysis and assumed complex analysis would be even worse. Boy was I wrong! As it turned out I loved complex analysis, and it has been my favorite branch of math ever since.

I thought the interval of convergence for ln(1-x) was -1 < x < 1. Does i fall in this interval so that we can use it in the series?

Can the Lambert Function be solved without a special calculator and how? I see lots of problems solved with the Lambert Function, but it strikes me as a piece of magic when it comes to finding the correct logarithm. What exactly is being multiplied or divided to get that answer?

Sir I have doubt if we open it using traditional method we got some standard solutions for summisions

Very nice.

How come you can’t just trig substitute x for (cotu)^2 and use trig identities to simplify the expression?

u = 1/x leads to int(1/(u²+1)²) and then u = tan(t) leads to int(cos²(t)) from 0 to π/2, which leads to the same answer of π/4

I tried it myself before I clicked and I thought I messed up somewhere, because when I checked my answer in Desmos it gave me 1.763... instead for some reason (floating point error?)

Sometimes I really hate the complex ones

Applying Lambert w function, I got that X=2.16 (approximately )

that's just how the constant is defined tho. you just "undiscretized" the summation into an integral using the floor function edit: nvm looked like you already said that in the video

I'm usually too lazy to do these but this was nice

Yes, I did in fact guess

Excellent

When I saw the thumbnail I suddenly had a craving for oily macaroni, as opposed to a quarter of a pie...

Nice Job!

Nice job!