Gaussian Integral with a (Reciprocal) Twist: Berkeley Integration Bee (1)

[@idavid8128]

You have to be at anime protagonist level of thinking full sentences in a second to be able to do this in 4 minutes

[@afreensadia2424]

Can you solve the problem?

[@adityadwivedi4412]

@@afreensadia2424 I can

[@afreensadia2424]

@@adityadwivedi4412 then give the solution

[@INSP_NITIN]

Nice video , calculus is the only part of maths which is still very useful for me , keep the good work up

[@hamiltonianpathondodecahed5236]

ay , didn't expect you here

[@dailyanimedoses6999]

_/\_ parnam sir ji

[@kaskilelr3]

I am the type of guy who virtually never hits the like button on youtube, but I found your channel and I've been liking every video I've seen so far. You are awesome, keep up the great videos.

[@typha]

The problem solving starts at around 0:40 and ends around 10:10. So it takes 9 minutes and 30 seconds. If you watch at 2x speed, that's 4 minutes 45 seconds... so if a person watches this video at 2x speed, and uses the right arrow key to skip ahead through predictable parts 9 times during the solution (which, granted, is a bit excessive), then they could have watched the solution in 4 minutes.

[@evanev7]

Stop watching me o.0

[@skrill500]

Lol squaring terms like (x^n+1/x^n) works out super nicely because you always have a constant term

[@slavinojunepri7648]

Excellent and inspiring solution to a tough integral at first glance.

[@plaustrarius]

Excellent! didn't go back for the second substitution and got stuck, really liked this thank you!

[@giovanicampos4120]

Boy I miss you please come back

[@ninadmunshi2879]

Hello! I was pleasantly surprised to see you uploaded this integral today. I'm actually quite impressed by your solution, I hadn't thought of using symmetry like this. The solution I had in mind does in fact use the same substitution you came up with! But the proof for it is a little different. We can assert that the integral of f(x) = integral of f(x-1/x) if the bounds are from -infinity to infinity. This is the Glasser master theorem en.wikipedia.org/wiki/Glasser%27s_master_theorem

[@sergeiivanov5739]

When any equality that involves two improper integral over R contains PV, does it imply that both of them might diverge, but when converge, then do so to an equal number, which is the Cauchy principal value?

[@ninadmunshi2879]

@@sergeiivanov5739 Yes that's right. Take the simple example of f(x) = x which is an example of your statement

[@nitayderei]

There is a nice sentence which simplifies that: integral from -infty to infty of f(x-1/x) equals to the original.

[@69erthx1138]

Just when I was thinking e^A+e^B=e^Ae^B, then splitting the integrations into a sum over {-infinity to 0}+{0 to infinity}....a Gaussian frown turned upside down.

[@Bemath_kh]

69erthx1138 Hasha

[@isaacaguilar5642]

This is a random fact but if u know it then it can be solved easily. Integral from 0 to infinity of f(x)dx give f(x) is even is the same as integral from 0 to infinity of f(x-1/x)dx. So if we write our integral as 2* e^-2 *int (0,inf) e^-(x-1/x)^2dx then its can be seen that its the same as 2*e^-2*int (0,inf) e^-x^2 which is 2e^-2 * sqrt(pi)/2 which is sqrt(pi)/e^2. The proof is basically what u showed but given a random f(x) thats even

[@nuranichandra2177]

Scary integral with an elegant solution. Great video

[@Bemath_kh]

Nurani Chandra soo scary

[@CipriValdezate]

So good... I had no words.

[@Celastrous]

Damn, really satisfying answer. Great video, very concise and easy to follow.

[@terryyoon1856]

My best friend in Berkeley actuslly sent me this!! Its pretty hard but ITS SO SATISFYING!

[@rajatkhandelwal7276]

Bro please start weekly maths challenges also

[@meiwinspoi5080]

Superb. The best. You rock.

[@biswadeepghosh5568]

Great solution 👍

[@tanusrimalakar9357]

Fantastic

[@stewartzayat7526]

THAT'S MY MAN!

[@pbj4184]

The reason the u-sub doesn't work is because of the singularity at x=0 which prevents the relation of u and x from being bijective

[@gamedepths4792]

Please continue the weekly maths challenge please!

[@mrl9418]

Bravo

[@lucas29476]

I wonder if this can be solved using contours?

[@marekkryspin8712]

I did it similarly (but more generally): There is equality ∫f(x)=∫f(x-¹/ₓ), when the integrals are at the limit -int to +inf. So if f(x)=e^-x², the left is √π and on the right is e² · (our integral) so our integral= √π/e².

[@gogolplex8576]

Cool😍 Do you know the name of that rule?

[@marekkryspin8712]

@@gogolplex8576 I did not know the name, but after a while of research I found: Glasser's master theorem (special case).

[@gogolplex8576]

@@marekkryspin8712 Ok, thank you very much! Apparently it is the Cauchy-Schlömilch relation, or at least it is related to it. Its always helpful to learn tricks like this, so thank you again for mentioning it

[@narekkousherian5015]

I did mine similarly too, solved with Gaussian error function

[@Bemath_kh]

All about tough integral🤯🤯

[@XanderGouws]

This is crazy. I like it.

[@yacinbadr5181]

ممكن الشرح باللغة العربية ؟

[@rishj007]

the problem just ends at step 3 when u come to know its an even function(exp. part).

[@rafael7696]

Great

[@prashantmishra9073]

It's easy with substitution logx=t

[@Smokie_bear9896]

pog

[@dileep_j203]

:) ❤❤👍👍

[@איתןגרינזייד]

I tried before watching the video and I almost got it and I am so mad

[@afreensadia2424]

Sir please solve the problem... If we can write a number as a^b then we call that number LAL number where both a, b is greater than 1. Again, the sum of two LAL numbers is a LAL number. After which value, every number is a LAL number?

[@hamiltonianpathondodecahed5236]

could you be more precise

[@afreensadia2424]

@@hamiltonianpathondodecahed5236 this is the actual question

[@hamiltonianpathondodecahed5236]

could you clarify the part which says ," _Again, the sum of two LAL numbers is a LAL number. After which value, very numberLAL number_ "

[@afreensadia2424]

@@hamiltonianpathondodecahed5236 like 9 is a LAL number, cause 3^2 = 9, again 16 is a LAL number cause 4^2 = 16. Now 9+16=25=5^2, 25 also a LAL number.

[@vishwanraja666]

finally something I can't do

[@shaswatadutta4451]

Could be solved within 3 minutes too.

[@wfe2891]

Go Golden Bears!! #Berkeley2024

[@zerospeed6412]

I literally managed to guess the answer in seconds.

[@lilyyy411]

:O

[@afreensadia2424]

Sir, please solve the problem... Suppose, Sheldon is gambling. There are numbers from 1 to 2n. He bid 666 taka for every even number and 3366 taka for every odd number. Each time a wheel is spinned with all those numbers. And a number is chosen randomly. Sheldon gets the money he bid on that number. After the first spin, how much should he earn?