we can also use the property of leibniz , \int^{\infty}_{0} \dfrac{x^k}{x^a+1} , here is pretty well known subs x^{\dfrac{a]{2}}= \tan(\theta) , then use beta function + euler reflection
I think we can use Contour integration on I(s) = int[R+] {x^s/x^a+1}dx And derive with respect to s evaluated at one, The preceeding mellin transform is evaluated as an exercise in gamelin’s complex analysis.
The huge swathes of videos with meaningless titles makes the task suggested @12:10 an exercise in futility. It's something I complained about on this channel for years before giving up as it seems to be something unimportant to the video presenter and editors.
They really should publish a register or kind of directory somewhere with Meaningful titles that serve as kind of a searchable table with links to those videos. 🧭
@@BikeArea There's already a register of the videos... it's the channel's video history. It is conceivable that a motivated individual could go through it and catalogue them. The issue by now is that there are thousands of videos. I thought about doing it myself in the early days (only a couple hundred videos at the time), but there'd be enough meaningfully titled ones that I let it go. Then he got grad students to work for him and I thought surely then they'd do it (I've been a grad student and know what kind of garbage tasks we get), especially with how deliberately malign the titles and video descriptions were. Then I tried to give up caring. (Unfortunately,) there's so much good stuff, it makes me angry that he's essentially writing text books and then throwing the pages into a blender, not caring if anyone bothers to piece them together.
The required identity can be found as the Fourier series of e^itx This video's result is related to dilogarithms, and the dilogarithms playlist could do with a few existing videos being added
@@xizar0rg I agree with you. It would be nice if there were an easier way to find the specific referenced videos, assuming it wasn't done by a completely unrelated 3rd party. Some day...
You get a power of pi times the m-th derivative of a cosecant, where m is the power of the logarithm. Not too hard to prove with induciton but I suggest you do the case 0,1,2 to understand what the general formula is. Hint: It's easier to call the power of x in the denominator "1/a".
I don't think the exercise makes sense. The initial inequality only holds true for x > 1, but for x < 1 the opposite holds true. But even if the inequality were true, the u*v term in the integration by parts doesn't converge either. lim(x->0) lnx/x^(a-1) -> -infinity/0, which isn't something L'Hopital's rule can handle.
I suppose we can't be too surprised. Inverse trigonometric functions are non-trivial logarithms of elementary functions. (By non-trivial, I mean they don't just cancel out. We can say x=ln(e^x), but that's asinine, hence trivial).
I'm not sure if this was the video he was thinking of, but this one is a derivation of the Mittag-Leffler pole expansion of sec(x) th-cam.com/video/eZD7uHlSWfc/w-d-xo.htmlsi=_PKRJ6lal5ia9b7q
The replacement x = 1/t remains valid throughout the derivation, so substituting again x for t may lead us to another substitution of x = 1/t which would be invalid.
we can also use the property of leibniz , \int^{\infty}_{0} \dfrac{x^k}{x^a+1} , here is pretty well known subs x^{\dfrac{a]{2}}= \tan(\theta) , then use beta function + euler reflection
I didn't understand a word you said. But I'm an engineer :-)
I don't think that bound at the beginning in the bottom left holds. Note that the limit at 0 doesn't converge.
this is such a nice integral, thank you
I think we can use Contour integration on I(s) = int[R+] {x^s/x^a+1}dx
And derive with respect to s evaluated at one,
The preceeding mellin transform is evaluated as an exercise in gamelin’s complex analysis.
13:46
It was indeed an undoubtedly a good place to stop.I couldn't have agreed more!
Combining the trig functions gives the answer -[ pi/a * cot(pi/a) ]^2 I believe
I think you would have to split that integral to (0,1) + (1,infinity) before doing your estimates.
The huge swathes of videos with meaningless titles makes the task suggested @12:10 an exercise in futility. It's something I complained about on this channel for years before giving up as it seems to be something unimportant to the video presenter and editors.
They really should publish a register or kind of directory somewhere with Meaningful titles that serve as kind of a searchable table with links to those videos. 🧭
@@BikeArea There's already a register of the videos... it's the channel's video history. It is conceivable that a motivated individual could go through it and catalogue them. The issue by now is that there are thousands of videos.
I thought about doing it myself in the early days (only a couple hundred videos at the time), but there'd be enough meaningfully titled ones that I let it go. Then he got grad students to work for him and I thought surely then they'd do it (I've been a grad student and know what kind of garbage tasks we get), especially with how deliberately malign the titles and video descriptions were.
Then I tried to give up caring. (Unfortunately,) there's so much good stuff, it makes me angry that he's essentially writing text books and then throwing the pages into a blender, not caring if anyone bothers to piece them together.
The required identity can be found as the Fourier series of e^itx This video's result is related to dilogarithms, and the dilogarithms playlist could do with a few existing videos being added
@@PyarMatKaro I appreciate what you've done for this specific instance. No cap, as the kids say. My point stands.
@@xizar0rg I agree with you. It would be nice if there were an easier way to find the specific referenced videos, assuming it wasn't done by a completely unrelated 3rd party. Some day...
Yes! I remember proving this very nice identity two years ago! I was also wondering what would happen when ln x was also raised to some power 🤔
You get a power of pi times the m-th derivative of a cosecant, where m is the power of the logarithm. Not too hard to prove with induciton but I suggest you do the case 0,1,2 to understand what the general formula is. Hint: It's easier to call the power of x in the denominator "1/a".
Same! For me I used the beta function and feymann's technique,
very nice!
Hi,
6:41 : missing dx .
But what is the reason for solving these equasions?
I don't think the exercise makes sense. The initial inequality only holds true for x > 1, but for x < 1 the opposite holds true.
But even if the inequality were true, the u*v term in the integration by parts doesn't converge either. lim(x->0) lnx/x^(a-1) -> -infinity/0, which isn't something L'Hopital's rule can handle.
Interesting result: if a goes to infinity, the integral approaches -1.
I suppose we can't be too surprised. Inverse trigonometric functions are non-trivial logarithms of elementary functions.
(By non-trivial, I mean they don't just cancel out. We can say x=ln(e^x), but that's asinine, hence trivial).
I'm not sure if this was the video he was thinking of, but this one is a derivation of the Mittag-Leffler pole expansion of sec(x)
th-cam.com/video/eZD7uHlSWfc/w-d-xo.htmlsi=_PKRJ6lal5ia9b7q
wow π shows up once again
The denominator of the integrand having an x^a+1 is generally a good sign of an inverse trig function, which often leads to pi.
The replacement x = 1/t remains valid throughout the derivation, so substituting again x for t may lead us to another substitution of x = 1/t which would be invalid.