a twist on a classic improper integral

[@applealvin9167]

actually it's not cos(7x) but cos(5x) also -f(0) which is -1 is missing so the answer should be (5/96)(25ln5-27ln3) (checked using desmos) btw this integral looks like dirichlet integral but there's no π in the final answer which is quite interesting

[@Jack_Callcott_AU]

@applealvin9167 You are right. He made a mistake writing cos(7*x) , and I checked on my HP50g calculator.👍

[@megauser8512]

But, it is *not* the Dirichlet integral, since it's integrand function is sin^5(x)/x^4, whereas the 5th Dirichlet integrand is sin^5(x) / x^5.

[@bobh6728]

dv=dx/x^4, not dv=dx/x^5

[@wolfmanjacksaid]

I was wondering why he got that antiderivative for v wrong, turns out it was right but he had dv written wrong as you point out

[@krisbrandenberger544]

That's right ✅.

[@Rócherz]

11:47

[@Jack_Callcott_AU]

Slight typo at 11:47 . It should be 1/x^4.

[@minamagdy4126]

Not mentioned in the video, but, early in the proof, the last term works because c_n = -(sum{i in [1,n-1]}(c_i))

[@coc235]

10:41

[@minamagdy4126]

@coc235 That is a different occurrence of the same fact coming up, that time more obviously and with a mention. I thought to clarify it for the earlier, less obvious time.

[@jayhem_klee]

5:00 seems that (c1+c2+..+cn)f(anx) is missing. Oh, it cancels nicely anyway !

[@megauser8512]

Exactly, since c1 + c2 + . . . + cn is assumed to be = 0.

[@talberger4305]

didn't you forget a -1 at 18:46 since you need to multiplied by -f(0)=-1. and also need to be ln(7) .

[@megauser8512]

Yes for the -1, but no for the ln(7) because ln(5) is correct, so he should've wrote cos(5x) on the above line, instead of cos(7x).

[@federicopagano6590]

I had written a theorem about this type of integral from 0 to imfinity of f(x)/x^n where the limit at the origin always exists as an hypothesis. Then this kins of integral can be computed =1/(n-1)! Integral [L(D^(n-1)f] ds from 0 to infinity where L stands for laplace transform and D regular n-1derivative. So in this case you have to take the third derivative of sin^5(x) because n=4 but before doing that you should expand sin^5(x) using the complex exponential definition of sin(x). Then you find laplace transforms which is very easy all cos(nx) involve and then you integrate all logarithms easy then you end up with same result (almost because he mixed signs of real result)

[@logician1234]

What is the motivation for that definition of f(∞)? It seems a bit arbitrary to differentiate between periodic and non-periodic functions.

[@bot24032]

the first definition doesn't work for periodic functions since the limit doesn't converge, and the second definition is the average value of the function in a sense. but i think the motivation for that would come from the fact that this definition works to find the Frullani integral

[@MrMctastics]

f(infinity) is whatever gets you the right answer

[@TheEternalVortex42]

It's basically just saying periodic functions become their "average" in the limit. Kind of like saying 1 - 1 + 1 - 1 + ... "converges" to 1/2

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@Hipeter1987]

This integral is actually Oops All Cosines 😁

[@nardmic]

in the third to last step it must be 25 cos(5x) and not 25 cos(7x)

[@Alan-zf2tt]

Aha! The monster is in the detail and once when detail is detailed the monster becomes zero leaving a remaining residue of some sort that does need to be worked out. How can a (devilish?) concatenation be identified from the start? Were I a brighter being I'd try to see if the complications of lots of zero results can be somehow bypassed by simpler tests if possible. But I realize my limitations and so as Michael says so frequently: this is a good place to stop?

[@CM63_France]

Hi, 8:38 : again a "question for you", may be Oscar will end his videos by "and that's a good place ..." we could have developed sin^5 x as a function of sin(kx) from the start, right?

[@r2k314]

Where does that definition of the value of the periodic function at infinity come from? How is it derived?

[@paolonicolacerea]

Am I wrong or all the a_i coefficients in the general Frullani integral should be positive? It is not specified. Thanks

[@tomholroyd7519]

Who says pictures of blackboards completely full of equations aren't real?

[@lucasf.v.n.4197]

does complex analysis solve all these improper integrals? (I mean residual theorem etc)

[@fartoxedm5638]

My approach to the question at the end of the video: We need to find such k that we have function sin^n(x) / x^(k+1) and d^k / dx^k (sin^n(x)) is a sum of sines(cosines) with coefficients sum to zero. First(thanks to Beyer, 1987, p 140) sin^(2n)(x) = 2^(-2n) * C(2n, n) + (-1)^n / 2^(2n - 1) * sum((-1)^k C(2n, k) cos(2(n-k)x), k, 0, n-1) sin^(2n+1)(x) = (-1)^n/4^n sum((-1)^k C(2n+1, k) sin((2n - 2k + 1)x, k, 0, n) Both of them are quite messy to work with so I'll proceed witht the odd one Sum of the coefficients of k'th derivative will be(I removed the constant factor part since it's already mess) sum((-1)^k C(2n + 1, k) *(2n - 2k + 1)^k), k, 0, n) and thanks to my several pages of combinatrics it's equal to zero when k is odd and less than 2n + 1 In the end we have that integral(sin^(2n+1)(x) / x^(k+1), 0, oo) can be solved exactly like in the video when k is odd and k < 2 n + 1 I'm kinda tired so you can derive the result for even power in reply

[@CTJ2619]

Why do you have do= dx/x^5 when in the original equation it is 1/x^4?

[@bobh6728]

I think he wrote that wrong. He found v by using x^4. So it didn’t affect the rest.

[@Patapom3]

Amazing!

[@r2k314]

Does some know link to original Frullani Integral video?

[@tutordave]

Using Derive 6, I got the negative of your final answer. Sign switched in there somewhere.

[@StanleyDevastating]

he doesn't evaluate the f(infinity) -f(0) term at the front, which equals -f(0) = -1

[@guntherbeer8234]

Why does one care about this? It has a name, so i would guess there's a reason one woukd care.

[@RexxSchneider]

It's troublesome that Michael Penn didn't spot that you can't get cos(7x) from any expression of the form (sinx)^n.(cosx)^m unless n+m is at least 7. I think it should be jarringly obvious that (sinx)^2.(cosx)^3 and (sinx)^4.(cosx) are going to produce at most cos(5x).

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x

[@federicopagano6590]

I wrote a theorem in which we can compute the integral of f(x)/x