Hardy's Integral

[@michaeljin101]

Thanks!

[@xinpingdonohoe3978]

@@michaeljin101 woah, you *really* wanted this integral done.

[@kostasch5686]

What an excellent approach. For anyone wondering, the simplest form using only factorials at the end is: In=pi/2*sum (2k)!*(2n-2k)!/[k!*(n-k)!]^2*a^(-1/2-k)*b^(-1/2-n+k)

[@PleegWat]

The final solution as displayed suggests it is negative half the time due to the (-1)^n term. However that cancels against minus signs embedded in the two binomial coefficients, and the value is actually always positive.

[@get2113]

Very clever. One of the professors best recently. My preference is analysis over number theory, but others might like all that prime number stuff.😅

[@whizgranny6203]

Inside Interesting Integrals... One of my favorite books!

[@EconAtheist]

Cool! I gotta get that book from the library.

[@goodplacetostop2973]

13:44 Where’s Laurel?

[@BarryRowlingsonBaz]

That's another fine math you got me into Stanley!

[@yoav613]

Laurel?

[@yoav613]

This laurel?kzbin.info/www/bejne/iZPPpaholKl0irMsi=RzrbJAYiWBReqfHt

[@BarryRowlingsonBaz]

@@yoav613 this Laurel kzbin.info/www/bejne/eGKWgqJ3ap6nd5Y because Hardy...

[@yoav613]

@@BarryRowlingsonBaz oh ok,thanks😃

[@calculuspro]

Nice integral

[@thatman3107]

Can someone help me understand the expansion using the binomial formula at 9:00?

[@jesusalej1]

Why make it easier if it can be made more complicated.

[@minamagdy4126]

I believe you can factor out a (-1)^n by turning the final combinations to be combinations of positive half-integers, whic would then cancel the same term outside the sum. Also, I would've liked to see what simplifications come about by taking a factorial-onlu representation

[@holyshit922]

My first Idea was to derive recurrence relation for I(n) but I tried to derive recurrence without using Leibnitz rule for integration

[@TedHopp]

Nice problem and solution. Maybe I'm being nit-picky, but at 13:40, it seems off the mark to call a summation "a nice closed-form solution". I would want to evaluate the sum in closed form before describing the solution that way.

[@BrianGriffin83]

So, after all, that was not such a good place to stop...

[@justcuzy3673]

True, it would be nice to have a more closed form, but here's what happens: Pull out constants- (1/2)!^2, a^(-1/2), and b^(-n-1/2) Remainder- (pi^2/2)*(-1/b)^n/(ab)^1/2xSum{1/k!*1/(n-k)!*1/(1/2-k)!*1/(1/2-n+k)!*(b/a)^k} One look at this sum should somewhat breakdown the idea; there is no closed form (no, I won't prove this here). As such, we leave the formula in its "cleanest" form, which is what was provided at the end. While it may seem unsatisfactory, there are many other such series definitions for the likes of the elliptic integrals and even definitions for transcendentals like pi and e.

[@TedHopp]

@@justcuzy3673 Oh, I have no problem with the solution as presented. I just wouldn't call it a "closed-form solution." I agree with you that the prospects for evaluating that sum in closed form seem rather dim.

[@get2113]

Give the prof a break. Hi final answer is simple to program up eliminating any need for numerical integration.

[@HPTopoG]

It’s a finite sum, so it counts as a closed form. Typically one only has issues with summations not being closed forms when they include infinitely many terms. (The main reason for not considering these closed forms is because of potential convergence issues.)

[@statebased]

Inside Interesting Integrals by Paul J. Nahin

[@yukfaicheung7484]

hardy, cambridge professor

[@Achill101]

Does the end result help us in other problems, compared to evaluating the integral atnthe start numerically for given n, a, and b?

[@Calcprof]

With the product of the binomials, factor out all the 1/2's, then you are left with two products of odd integers, which you re-write in terms of factorials. {\displaystyle (2k-1)!!={\frac {(2k)!}{2^{k}k!}}={\frac {(2k-1)!}{2^{k-1}(k-1)!}}\,.} I'm not sure this makes things simpler, because I haven't actually done it! 🙂

[@GokmatematikMATEMatik]

HEY AM HARDYY HE I AM EASLYYYYYYYYYYYYY

[@mohamedtahiri4388]

It's very important and higher technics of solving this type of integral thanks for sharing this

[@theartisticactuary]

I'd tidy it up a little more. Bring b^n and 1/sqrt(ab) common factors out to where they belong, to the left of the summation sign

[@josepherhardt164]

Do mathematicians lie awake nights composing "impossible" integrals?

[@douglaszare1215]

Typically at the time of Hardy, physicists or engineers would encounter integrals like this when performing some approximation (suppose cows are not perfect spheres, but oblate spheroids--how does that perturb the results?) or an inverse Fourier transform, and when they had trouble they would ask a mathematician for help.

[@dukeofvoid6483]

Coding a function to numerically solve that to any accuracy is trivial, so no they don't - if they're of sound mind.

[@AnkhArcRod]

Can we please all agree to just call it Leibnitz rule and not Feynman's trick?

[@primenumberbuster404]

What's in the name? We call it the "try differentiating under the integral" rule.

[@debtanaysarkar9744]

Nooooooo

[@davidruhdorfer3857]

I would agree, but the guy's name is Leibniz, not Leibnitz.

[@bjornfeuerbacher5514]

@AnkhArcRod: Leibniz invented the rule, but as far as I know, it was Feynman who first used it extensively for actually calculating integrals. So both names fit.

[@bjornfeuerbacher5514]

@@davidruhdorfer3857 That's true starting from 1671; before, there were varying different versions of how his name was written. (And his father's name was actually Leibnütz.)

[@kmlhll2656]

Waw waw waw !

[@seanfife]

Any way this extends to Rational numbers for n?

[@aidansgarlato9347]

can you analytically continue the last statement for n to the complex plane?

[@TomFarrell-p9z]

I was wondering about whether the technique could be modified for real, or at least rational n. But then there is no termination for the iteration of integrals where Michael solves for n = 1.

[@Alan-zf2tt]

Knee jerk reaction at 0:02 Oh no! It has got an I in it. A capital I. And it is followed by 'n' a lower case 'n'. Two tricky letters added together must make a very tricky integral In 🙂 in 'I' and in 'n'

[@hoodedR]

Could have all been a lot simplier bh exploiting symmetry at any step before he binomial expansion. we can see that I is symmetric about the transformation ab and so dI/da = dI/db. You can also prove it rigorously by using the substitution x=π/2-x in either expression for dI/da or dI/db

[@alricboullemant3117]

The substitution doesn't quite work, you'll have an a*sin(x)^2 + b*cos(x)^2 in the denominator, so dI/da ≠ dI/db... maybe you meant that we have dI/da (a,b) = dI/db (b,a) and vice-versa, which I'm pretty sure is definitely the case. This doesn't really simplify stuff tho If there was that first symmetry then the final result would indeed have a much nicer expression

[@Achill101]

Penn used the symmetry by writing the forms for b without discussion, because it was the same form as for a.

[@hoodedR]

@@alricboullemant3117 ahh yeah you're right 🤔 I jumped the gun there a bit. My bad

[@gp-ht7ug]

From 09:04 onwards 🤔😵‍💫

[@MrMctastics]

One short class in multiplicative number theory or enumerative combinatorics will make you sick of discrete convolutions!

[@shubhanyujain5208]

first