Decrease That Exponent: Berkeley Integration Bee (2)

[@ninadmunshi2879]

This was actually used as the tie breaker for the finals round! The contestants had 5 minutes to do this, but the winner finished much more quickly. Beautiful explanation as always. A fun solution is to use Feynman trick to get the 2020th derivative of int x^n dx from 0 to 1, which can be obtained from geometric series.

[@duggydo]

This is very interesting. An integral that can be used to represent factorials. Very nice!

[@geometrydashmega238]

Check Euler's gamma function. This integral can be expressed as such by the substitution lnx = -u, where you get gamma(2021) which is 2020!. Just Google Euler's gamma function if you're interested.

[@virajagr]

@@geometrydashmega238 this thing is bugging me for a long time. Why is gamma(x) = (x-1)! ? Why not simply gamma(x) = x! ? Why define it the complicated way?

[@benjaminrosenberg]

@@virajagr I think the idea behind the gamma function being (n-1)! is that the integrand is x^(n-1)e^(x) so you can look at the integral of say, x^(4)e^(-x) and realize immediately that it's 4!. There might be another more complex reason than this but that's what came to mind for me

[@geometrydashmega238]

@@virajagr Unfortunately I can't help you on that. I've searched and it seems more like a historical issue (for whatever reason someone preferred this gamma), but some people give arguably interesting reasons for it, which I didn't understand because my math level doesn't reach that. If you are interested google 'why is gamma function shifted by 1' as I did, and maybe check the Stack Exchange discussions.

[@geometrydashmega238]

@@benjaminrosenberg He wasn't referring to that, but to the actual definition of gamma(x). He is asking why wouldn't you better define gamma(x)= integral from 0 to infty of t^x * e^(-t) dt instead of gamma(x)= integral from 0 to infty of t^(x-1) * e^(-t) dt (see the change between x and x-1). The first one which Viraj thinks as the 'natural' gamma definition is known as pi function, but isn't widely used.

[@idavid8128]

Very cool solution! Another interesting way to solve this one is to let -u = logx, from that and making some observations the integral definition of factorial numbers pops out ! That's the way I solved it

[@virajagr]

Same, that gave us the gamma function

[@axemenace6637]

Wow. That's a really, really fast solution - you could easily do this in under 30 seconds!

[@Ty-sc2ri]

Why not just let t=-ln(x)?

[@97kerala]

Yup, I did this and this simplified the problem considerably! (Only thing is you need to be allowed to do is use the gamma function integral expression)

[@rhythmmandal3377]

@@97kerala Even if you couldn't use the gamma function you could show in like a 1 min that the function is 2020!.

[@sunitajha7596]

The integral screams *Gamma Function.*😂😂

[@sunitajha7596]

Just plug in x = exp(-u) and it spits out 2020!

[@alexting827]

@@sunitajha7596 In fact, solving the general case for k instead of 2020 gives (-1)^k * k! It's very cool!

[@depressedguy9467]

One can use fayman trick here use integration from 0 to 1 x^a dx Now differentiate it w.r.t a u will get x^a log x Repeat this 2020 time so u will get x^a( logx)^2020 Since above original integral has ans 1/(a+1) Differentiate it 2020 time u get 2021 !/(a+1)^2021 Put a=0 u get 2021 factorial Another way is using gamma function substitute u= -logx and solve it yourself

[@aniruddhvasishta8334]

He says "let's dive right in" at around 0:33 (just to make the math easy) and the video is 4:53 minutes long so that means he solved in 4:20, nice!

[@shaswatadutta4451]

A brother of the gamma function

[@Dionisi0]

What happened to you?

[@particleonazock2246]

You must be Korean, with that strong Korean accent, which really adds flavour to your videos.

[@solitary6002]

Substitute u = ln x. Bounds become -∞ to 0. The integral becomes u^2020 • e^u du Now use integration by parts to get a series: e^u•( u^2020 - 2020•u^2019 + 2020•2019•u^2018 - .... + 2020!) For u = 0, e^u = 1 and all ther terms become zero except 2020! For u=-∞, e^u = 0. So answer is 2020!

[@Niglet101bitch]

YAY! That was the fist Integration Bee problem, which I was able to solve on my own. Thank you for the cool video. :)

[@natevanderw]

Just found your channell. It looks like you stopped making video 7 months ago, but I hope you continue to make some more content.

[@bryanmoreno8925]

I love your videos!!

[@mab9316]

Sir, What the name of software you are using for writing? Thanks.

[@giovanicampos4120]

My booooy I do miss you 😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭

[@Munnu-hs6rk]

Take some jee advance problems you will love it

[@vishalstyles161]

Can you please make videos on complex numbers (advance level )?

[@ucchashhore3353]

can anyone provide me the solution with gamma function?

[@alexting827]

Fun fact, I actually came up with the general question integral from 0 to 1 of ln^k (x) on my own time and it's possible to derive the general case as (-1)^k * k! using the gamma function! I had this memorized so if I were at Berkeley this would have been an instant answer LOL

[@amanattheedge9056]

can also be solved with gamma function

[@rhythmmandal3377]

took me 30s to realize this was gamma(2021)

[@Fizington]

Another brilliant video! What software do you use for writing these solutions?

[@angelmendez-rivera351]

Not surprising that this integral representation can be used to express (-1)^2020·2020!, since the original integral definition for the Gamma function given was the integral (0, 1) on [-ln(x)]^z with respect to x. Eventually, this was replaced by the exponential form, because it was more convenient that way, and for historical reasons, a cumbersome shift was introduced, so that x^(z - 1)·exp(-x) became the new integrand on the interval (0, ♾).

[@desmondhutchinson6095]

Does anyone know where a complete list of the integrals presented during the integration bee are? I cant seem to find many, just portions of videos on youtube.

[@ninadmunshi2879]

Hello! I wrote the problems for the bee. While I can't provide all of the problems that were given during the live competition, I can send you the qualifying test and answers. The integrals were a mixed bag of easier and harder than the live bee integrals.

[@desmondhutchinson6095]

Ninad Munshi Please do!

[@themaverick1891]

@@ninadmunshi2879 Could you send it to me on email? I'll give you my email address.

[@ninadmunshi2879]

If both of you can email me, that would be great

[@desmondhutchinson6095]

Ninad Munshi [e-mail] is my email.

[@RomanoPRODUCTION]

Beautiful

[@vishwanraja666]

beautiful

[@림林]

I am not good at English but I could understand well, it's amazing

[@chessandmathguy]

Beautiful.

[@nithin1729s]

I first tried to derive the reduction formula to this integral.

[@xaxuser5033]

unbelievable!!!!!! try some functional equations problems they r soo freaking good

[@maxsch.6555]

I just defined I(t) to be the integral from 0 to 1 of x^t dx. It' easy to see that the 2020th derivative of I(t) evaluated at t = 0 is our integral we want to find. On the other hand is I(t) = x^(t+1)/(t+1) |₀¹ = 1/(t+1) The nth derivative of 1/(t+1) is (-1)^n * n! (1/(t+1))^(n+1). So the the 2020th derivative of I(t) evaluated at 0, our wanted integral, is (-1)^2020 * 2020! * (1)^2021 = 2020!

[@MaxxTosh]

What’s your favorite meganumber? Is it 2020!?

[@TechToppers]

Where are you? There are 9² comments on the video.

[@cantcommute]

Working hard at Harvard lol. We'll probably see him next summer tho can't wait for his return!

[@mohan153doshi]

Please switch over to the DI method for Integration by parts.

[@easymathematik]

For those who are looking for some challenges involving polynomials: "Are you tired of polynomial long division?" -> kzbin.info/www/bejne/r2WTpJmVpq9jipI Enjoy.

[@JuanLopez-rl7ry]

You can make this more general: int_0^1 (ln(x))^z dx; x = exp(-y); dx = -exp(-y) dy; y_bottom = ∞; y_top = 0; therefore int_∞^0 ((ln(exp(-y)))^z)*exp(-y)*(-1) dy = int_0^∞ ((-y)^z)*exp(-y) dy = int_0^∞ ((-1)^z)*(y^z)*exp(-y) dy = ((-1)^z) * int_0^∞ (y^z)*exp(-y) dy = ((-1)^z)*gamma(z + 1) = ((-1)^z)*(z!)

[@mokouf3]

Assume that you are allowed to use gamma function u = ln(x) x = e^u dx = e^u du Write u as function of x: lim(x→0+) u(x) = -∞ u(1) = 0 ∫(ln(x))^n dx (x from 0 to 1) = ∫u^n e^u du (u from -∞ to 0) t = -u, dt = -du ∫u^n e^u du (u from -∞ to 0) = -∫(-t)^n e^-t dt (t from ∞ to 0) = (-1)^n ∫t^n e^-t dt (t from 0 to ∞) = (-1)^n Γ(n+1) Since n = 2020 which is even integer in this case ∫(ln(x))^n dx (x from 0 to 1) = (-1)^2020 Γ(2021) = 2020! Even if you are not allowed to use gamma function, you can use DI method to speed up the integration by parts.

[@shivamkushwaha1061]

Looks like he stopped posting videos feeling sad😢

[@targetiitbcse1761]

maybe due to less number of views and likes. KZbinrs get less support initially and that's sad

[@jaykay2218]

Another 2020 monstrosity

[@lambdamax]

This is lit. Thank you.

[@ehteshamashraf7627]

Area = 3.86 × 10^5802 unit²

[@isaacaguilar5642]

The integral from 0 to 1 of (lnx)^t dx = t!

[@JuanLopez-rl7ry]

Nope, the correct answer would be ((-1)^t)*gamma(t + 1) = ((-1)^t)*(t!) You are promoting heresy right now

[@angelmendez-rivera351]

The integral on (0, 1) of [-ln(x)]^t with respect to x would be equal to t!. What you commented is (-1)^t·t! instead.

[@isaacaguilar5642]

Angel Mendez-Rivera i mean to add if t is even which was an assumption i made based on this problem

[@ryandx8088]

logx is different lnx

[@ramk4004]

Great

[@user-zb4lu9rc6z]

why du=dx ,anybody

[@MA-bm9jz]

Much easier to do x=e^u

[@whatthehellisthis6245]

Yes, his method is more complicated

[@supriyamanna8412]

love from india bro.......

[@sunritroykarmakar4406]

pretty easy..use recursive relation of log^n x dx

[@thunderstorm178]

*gamma function*

[@isaacaguilar5642]

Solved it in under a minute with -u=lnx and knowing the gamma function.