First for first time lol. I have a cool integral for you: int (0,infinity) ln²(x)/((√x)(1+x²))dx

@XclusiveScienceSecrets

25 күн бұрын

3*pi^3/2^(5/2)=16,4435....

@aleksandervadla9881

24 күн бұрын

Isnt it π^3 / 8

@XclusiveScienceSecrets

24 күн бұрын

@@aleksandervadla9881 No. You can trust me, because I verify my results by numerical integration. About a month ago Maths_505 considered a similar integral with ln(x)^2 in the numerator. The solution reduces to calculating the second derivative of some cosecant at point 0. This method is quite applicable to this case as well.

There are many hidden identities that can be simplified further. For example Γ(1/6)Γ(5/6) = 2π Γ(1/3)Γ(2/3) = 2π/√3. Similarly, writing Γ(4/3) = Γ(1/3)/3 and Γ(7/6)=Γ(1/6)/6 you can simplify the other set of fractions. Not sure everything eventually comes to a nice form though.

@GreenMeansGOF

25 күн бұрын

The final answer can be written in terms of only Gamma(1/3) but it doesn’t seem to simplify further.

@dogsteady618

24 күн бұрын

The final result is like 1 - someting * π^2 / Γ(1/3)^3 + something * Γ(1/3)^3 / π, per the reflection formula and duplication formula.

@emma5068

23 күн бұрын

The solution is pretty messy but it's technically simplier in that it uses less of the gamma function and more elementary functions instead. Fully simplified the solution is: 1 - (2 × 2^(1/3) (sqrt(3) - 1) π^2)/(3 Γ(1/3)^3) + ((2 sqrt(3) - 3) Γ(1/3)^3)/(12 × 2^(1/3) π)

@ericthegreat7805

23 күн бұрын

Eulers reflection formula right?

@emma5068

22 күн бұрын

@@ericthegreat7805 Of course.

This is stunning. Thank you very much.

Alhamdulillah I came out alive...😂😂😂 You gave me a good laugh. After all these sophisticated integrals that you solve so elegantly... It's like Einstein scratching his head to find the answer to 1+1 🤣🤣

Do you think that you could do a video explaining that initial phase shift, specifically where it is and isn't valid?

Very nice!.i tried to solve it first,i failed,so i looked for the result at the end of the video,and when i saw it i realized that i must watch the video😂💯💪

I was able to solve this on my own. nice

OKAY COOOOOLL!!!!!

Hey! I want to get into writing on a tablet like you, but I haven‘t found a nice app for that. May I ask which one you use?

very cool integral and yeah, fractions are the Absolute Monsters of maths 😢😂

Itd be interesting te generalise it to the nth root

check the values for u and v in beta

@BridgeBum

24 күн бұрын

U and v are backwards, but I don't know for certain but is there symmetry? That is, is B(u,v)=B(v,u)?

ok cool !!

OpenAI provided result Pi/4 including nice calculation details

we know that the integral of sin(x) is equal to the integral of cos(x) if we go from 0 to PI/2, and therefore we substitute, but if we go from 0 to another number, and therefore the integrals are not equivalent, what we do?

Hello, I am curious, what application are you using to write this integral out on?

@user-rv2wj7pn5m

25 күн бұрын

Samsung notes

a result with 6 times the gamma function can only be beautiful...

@EtienneSturm1

23 күн бұрын

Also you can express gamma of 2/3, of 4/3, of 5/6 and of 7/6 as functions of gamma of 1/6 and of 1/3 and reduce further the result to something more elegant

Hi, May be int_0^{pi/2} f(sin x) dx is always equal to int_0^{pi/2} f(cos x) dx , no ? "terribly sorry about taht" : 0:41 , 2:40 , 5:39 , 7:55 , "ok, cool" : 5:27 , 7:08 , 9:35 , 10:34 .

@maths_505

22 күн бұрын

Yes it is always equal my friend.

Surely that answer simplifies. Please make a video on it :)

@maths_505

25 күн бұрын

No way it looks so cute as it is🥺

@deweiter

25 күн бұрын

Actually, all we have to do is to find closed form for Г(1+x)*Г(y)+Г(1-x)*Г(1-y). After that we just have to do some arithmetic, plug values for x and y, and lastly use the Г(x)*Г(1-x) formula

7:04 the integral defination of the beta function is always defined when the values are striclty non negitive . The form which is obtained on anylitical continuation should not be taken with integral notation ig . Please help

@SussySusan-lf6fk

22 күн бұрын

You're right. That's why I did it in a different way.

@venkatamarutiramtarigoppul2078

19 күн бұрын

@@SussySusan-lf6fkhow

Mathematica gets this as 1+Sqrt[\[Pi]] (-(Gamma[7/6]/Gamma[2/3])+Gamma[4/3]/Gamma[5/6]+Gamma[5/3]/(4 Gamma[7/6])-(2 Gamma[11/6])/(5 Gamma[4/3]))

0:31 The limits of integration ...?

@maths_505

25 күн бұрын

Are exactly the same

@davide5420

25 күн бұрын

Aren't they the opposite?

@neilgerace355

25 күн бұрын

​@@maths_5050 goes to π/2 and vice versa, then to flip them back, you need a minus sign. What have I missed?

@maths_505

25 күн бұрын

@@neilgerace355 the dx also transforms into -dx so the 2 negatives cancel out.

@neilgerace355

25 күн бұрын

@@maths_505 that's what I missed :) let me know if you need any help combining fractions :)

I evaluated it in different way and got more beautiful result. This integral is exactly equal to pi/4

@princeyadav1233

23 күн бұрын

I can explain if someone replies to this comment

@XclusiveScienceSecrets

23 күн бұрын

@@princeyadav1233 The value of this integral can be obtained by numerical integration. It is equal to 0.872207507697+-10^(-12). The value of pi/4 is 0.785398163397... So you are wrong.

Can you/anyone help me with this question ,I just can't find it's solution anywhere,this que was on my test.Pls help me Int(0 to 1) (xln²x/(1-x⁴))=k(1+1/3³+1/5³...infinity) Find k This has to be solved only using elementary methods(because I am in 12th rn)u can use Taylor expansion if it helps

@XclusiveScienceSecrets

25 күн бұрын

The indefinite integral is equal to (2*ln(x)*ln(x^2+1)+Li_2(-x^2))/4+C The definite integral is equal to -pi^2/48 The sum of the series is equal to 1/2*(Li_3(1)+eta(3)) All the theory you need is contained in the Wikipedia article “Polylogarithm”

@syed3344

25 күн бұрын

@@XclusiveScienceSecrets that doesn't answer my question, I asked rhe value of k

@maths_505

25 күн бұрын

Expand 1/(x^2+1) as a geometric series and the only other tool you'll need is integration by parts.

@syed3344

25 күн бұрын

@@maths_505 sorry i had written the que incorrect ,I have edited it now

@maths_505

24 күн бұрын

@@syed3344 then expand 1/(x^4-1) as a geometric series.

🗿

"Idk how I made it out alive" 😂😂😂 "I'm not very good at math" I notice as you get into more advanced math with greek letters you forget how to do basic arithmetic with numbers 😂😂😂 Edit: Btw the denominators look like you are supposed to multiply them and use eulers reflection formula to simplify the fractions. 1/6 and 5/6, 2/3 and 1/3.

@maths_505

23 күн бұрын

Yeah but they were so damn cool so I didn't want to disturb the gammas 😂😂