You need to try this integral

Evaluating a super HARD integral with an incredible shortcut: Cauchy-Schlömilch transformation.

[blackpenredpen]

Thanks to Avraham Radin for pointing out my mistake in the previous upload.


"56 minutes ago
At 11:27 it's missing the 1 over a"

[pointlesssentience]

For the last question: divide the numerator and denominator by x^2 and complete the square in the denominator to get 1/((x-1/x)^2 +3). Applying the transform and integrating to get an arctan expression.
Answer is pi/(2sqrt3)

[AdiBenIsrael]

It is interesting that the result does not depend on the constant b, assumed positive.

[technoguyx]

Imagine how crazy Cauchy must've been to discover this back in his day. :P Nice video as usual, your proof of the identity was very clear.

[fonaimartin]

Subbing -x and adding it to the original integral yields twice the original integral on the LHS and 2cosh(x)exp(-sinh²(x)) on the RHS, and another sub reduces it down to the integral of exp(-x^2) on the whole number line, which is well known.

[jimnewton]

I've never heard of the Cauchy-Schlömilch transformation. That's pretty cool in fact. Reminds me a little bit of the Euler substitution, which I think you've done another video on. Nice technique.

[mohandoshi]

Fantastic evaluation and an equally fantastic explanation. Please do some more videos on the CS Transformation.

[jerryjin]

Pease make more videos on famous integrals like this, love them, many thanks!

[andabata]

There is a direct way to evaluate the integral. Note, by substituting x -> (-x) that I = Integral[Exp[x-(Sinh[x])^2], {x, -Inf, Inf}], = Integral[Exp[(-x)-(Sinh[x])^2], {x, -Inf, Inf}]. Adding the two gives 2*I = Integral[(Exp[x-(Sinh[x])^2)+Exp[(-x)-(Sinh[x])^2], {x, -Inf, Inf}] =Integral[(Exp[x]+Exp[-x])*Exp[-(Sinh[x])^2], {x, -Inf, Inf}]. Dividing by 2 and using the definition of Cosh, we have I = Integral[Cosh[x]*Exp[-(Sinh[x])^2], {x, -Inf, Inf}]. Substituting u = Sinh[x] gives I = Integral[Exp[-u^2], {u, -Inf, Inf}] = Sqrt[Pi].

[yoshikagekira]

The CST is very incredible. Thank you for the video

[VibingMath]

wow what a nice proof of cauchy transformation man! And the heat question is also a lot of fun!

[fengshengqin]

Let ax-bx^(-1)=u, then dx=du/2a +udu/sqr(u^2+4ab), then the integral can be divided into two integral in the u world, from -infinity to the +infinity. the first part is integral 1/2a f(u^2)du, the second part is u/sqr(u^2+4ab) times 1/2a f(u^2)du . Well the second part integral is equal to 0, because this function is odd function and integrated from -infinity to +infinity. the fisrt part integral is a even function, so it can be written into the final result .

[dlevi]

Box at 16:19 shows the wrong integration limits (should be from -inf to +inf)

And just confuse the hell out of everybody and write the result as Γ(1/2)

[happyhe]

Hey blackpenredpen, I recently saw two integral problems that I think you could give a neat explanation to it. It goes like this :
1.integral dx from 0 to ∞

2.integral 1/(x+(1+x^2)^(1/2))^2 dx from 0 to ∞

[abhirupchakraborty]

Can you suggest a good book for calculus where we can find such beautiful transformations and formulae? In the regular books there are no such useful tricks. Hence, if you could suggest any book, that will be very much helpful for us.

Thanks for such videos. These enrich us a lot. 🙏🙏🙏

Best wishes.... From India

[marinmaths]

Clever substitution! I'm gonna try the problem for later...

[carlosgiovanardi]

Thanks for sharing! Big trick! And also great teacher!

[forgetfulfunctor]

CUTE PROOF, HAVEN'T SEEN THIS TRANSFORM BEFORE. NICE VID

[WhattheHectogon]

7:17 What the heck...togon!


fixed that for you <3

[thephysicistcuber]

I was literally thinking of something like this to do the integral of exp(ax^2+b/x^2) from 0 to inf a few hours ago. Then I come across this video. Lol.