Back with the Fresnel integrals but this time we're evaluating them using an approach you've probably never seen before: using the wonderful gamma function.
At last A reasonable solution that I can understand I have been trying to know a know how to do these for months but since I am a high school student most solutions were either or too hard for me(since when I learn something I want to know everything from the scratch) I understood everything except your solution of the √tan(x) integral Luckily you posted a video 2 months ago with a simpler solution Thank you for the video
I was looking at solutions of the Dirichlet integral on an old Math SE page after watching this video and someone mentioned using what's called the Schwinger parameterization: en.m.wikipedia.org/wiki/Schwinger_parametrization. As soon as I clicked on the Wikipedia page, I realized you used exactly that to evaluate this integral. Awesome solution! I'll definitely see if I can apply this technique on future integrals.
Does anyone remember the Gaussian integral? 0 to infinity integration of e ^(-ax^2) so -a=i we know that this integral is equal to 1/2 x sqrt(pi/a) so put a=-i we get our result to be 1/2 x sqrt (pi/(-i))= 1/2 x sqrt(ipi) = [sqrt (pi)/2 ] x sqrt(i) and we all know what is square root of i
Great indirect method to solve these innocent-looking integrals. Just one question, when cot was turned to tan, isn't there a missing minus sign due to the reversion of the boundaries? I must be mistaken as the result seems correct, but can't see how. Thank you.
Thanks for the hard work. BTW, I don't understand the xt substitution. It looks like the integral expression on 3:33 under variable u is the same as the expression under t on 2:58. That would imply that 1/x^1/2 had to be equal to one which it is obviously not the case. Am I missing something? Again, thanks.
I would approach this buy integrating exp(i x^2) over a limiting contour which in this case would be a sector of radius R>0 and angle pi/4, let R tend to infinity so that the contribution from the circular arc goes to zero
Oh, I've got fond memories of this one. Back in high school I tried to do it without complex analysis for some reason (probably a stupid bet over a drink) and ended up spending a week on it. In the end the solution ended up with another well-known integral of dx/(x⁴ + 1), which incidentally had been taught in that very same course...
@@illumexhisoka6181 Well, you start off routinely and substitute x²= t, then think very hard about Γ(1/2). Having a drink at this point helps. Factorise out Γ(1/2) in the denominator and you're left with something like dtdϕ(sinϕ.e^-tϕ)/sqrt(ϕ) inside a double integral. Getting rid of the t-integral (second drink optional) will get you ϕ^-½/(1+ϕ²) at which you substitute x'= sqrt(ϕ), et voilà.
I heard it slightly involved complex numbers, ran as fast as I can. 11:18 well you know what they say, "small mistakes makes rockets crash to planets, or a heap of math nerds really angry."
I tried to come up with a general formula for the power is n(any positive integer) instead of 2 using your method But I stopped at the last step Integrate tan^a(x) from 0 to π÷2 Is there an easy way to do this?
Try either the beta function or residues If you're not comfortable with the beta function, watch my video on the michael penn integral (and share it too plz!)
@@maths_505 it took me a while to find a simple video that proof the relationship between the gamma and the Betta functions but I fainaly found it The integral from 0 to ∞ of e^(i(x^a))(when a is any positive number greater than 1) is equal to ((Factorial((-1/(2*n)))*Factorial((((-2*n)+1)/(2*n))))+((I*Factorial((((-n)-1)/(2*n))))*Factorial(((1-n)/(2*n)))))/((2*n)*Factorial((-1/n))) Not the simplest way to write it but this is the best I could I wrote it in terms of the factorial function instead of the gamma function because it Confuses me sometimes It surprised me that if the power changed the values of the two integrals won't be the same any more I am not sure if the formula work with complex numbers When I substituted with I in the formula it gave me a finite value but when I computed the integral in Wolfram alpha to compare the answers it didn't give me anything Again thank you for the video
Another way to do it, its by studying I(t)=(integral from zero to infinity)exp(-tx^2) the feynmann way. Then you substitute t=i and you evaluate the real and imaginary parts
Great solution! I personally would have to liked it if you'd taken a little bit of time to appreciate the fact that the integral of the sqrt(tanx) is in fact equal to its reciprocal over the interval from 0 to π/2, something which has been astounding for me ever since I found it out! It's crazy!!!
Question: when you evaluate the limits, you say that when x->infinity, exp(-x*(t-i)) goes to zero. The outer integral w.r.t. t goes from 0 to infinity. The limit above does not converge when t=0. Is it ok to say the case where t=0 is excluded in the outer integral because you can always remove null-sets from integration domains and still get the same result, i.e. the integral over t>=0 and t>0 are the same?
You didn’t have to do all of that. Suppose that the parametrised gamma function integral is denoted as J. Which you call I_1 (i will call it J). The value of J(x) = sqrt(π) / sqrt(x) Our original intgral I is just 1/2 ( J(i) ) We can just substitute x = i in the above formula and get the answer! There is no need to do a double integral other than to flaunt you integration skills of course!
I've seen you use this trick of finding the representation of a function in some different way, and maybe you know this but im surprised you haven't pointed it out, its just a Laplace transform. To illustrate, maybe you can do one integral you did a while back with this method, the int from 0 to inf of 1/(1+x^4) dx. i.e find a representation of 1/1+x^4 in terms of a certain laplace transforms and use it to get the answer, pretty neat stuff.
it took me a while to find a simple video that proof the relationship between the gamma and the Betta functions but I fainaly found it The integral from 0 to ∞ of e^(i(x^a))(when a is any positive number greater than 1) is equal to ((Factorial((-1/(2*n)))*Factorial((((-2*n)+1)/(2*n))))+((I*Factorial((((-n)-1)/(2*n))))*Factorial(((1-n)/(2*n)))))/((2*n)*Factorial((-1/n))) Not the simplest way to write it but this is the best I could I wrote it in terms of the factorial function instead of the gamma function because it Confuses me sometimes It surprised me that if the power changed the values of the two integrals won't be the same any more I am not sure if the formula work with complex numbers When I substituted with I in the formula it gave me a finite value but when I computed the integral in Wolfram alpha to compare the answers it didn't give me anything Again thank you for the video
At last
A reasonable solution that I can understand
I have been trying to know a know how to do these for months but since I am a high school student most solutions were either or too hard for me(since when I learn something I want to know everything from the scratch)
I understood everything except your solution of the √tan(x) integral
Luckily you posted a video 2 months ago with a simpler solution
Thank you for the video
Good use of the Beta and Gamma integrals. Impressive choice of substitutions and understanding of the relationships between these tools.
I was looking at solutions of the Dirichlet integral on an old Math SE page after watching this video and someone mentioned using what's called the Schwinger parameterization: en.m.wikipedia.org/wiki/Schwinger_parametrization. As soon as I clicked on the Wikipedia page, I realized you used exactly that to evaluate this integral. Awesome solution! I'll definitely see if I can apply this technique on future integrals.
Special solution , beautiful .
It will be easier to understand if you can write out the prerequisite knowledge you will use at the beginning.
Does anyone remember the Gaussian integral?
0 to infinity integration of e ^(-ax^2)
so -a=i
we know that this integral is equal to 1/2 x sqrt(pi/a)
so put a=-i
we get our result to be 1/2 x sqrt (pi/(-i))= 1/2 x sqrt(ipi) = [sqrt (pi)/2 ] x sqrt(i) and we all know what is square root of i
nice approach
adventurous approach
Great indirect method to solve these innocent-looking integrals. Just one question, when cot was turned to tan, isn't there a missing minus sign due to the reversion of the boundaries? I must be mistaken as the result seems correct, but can't see how. Thank you.
There's also a negative sign with the differential element which cancels out the other negative sign due to reversion.
@@maths_505 oh ok now i see.
Keep up the good work
Thanks for the hard work. BTW, I don't understand the xt substitution. It looks like the integral expression on 3:33 under variable u is the same as the expression under t on 2:58. That would imply that 1/x^1/2 had to be equal to one which it is obviously not the case. Am I missing something? Again, thanks.
New sub. Appreciate your work.
I would approach this buy integrating exp(i x^2) over a limiting contour which in this case would be a sector of radius R>0 and angle pi/4, let R tend to infinity so that the contribution from the circular arc goes to zero
Oh, I've got fond memories of this one. Back in high school I tried to do it without complex analysis for some reason (probably a stupid bet over a drink) and ended up spending a week on it. In the end the solution ended up with another well-known integral of dx/(x⁴ + 1), which incidentally had been taught in that very same course...
How did you go from these integrals to 1/(1+x^4)
@@illumexhisoka6181 Well, you start off routinely and substitute x²= t, then think very hard about Γ(1/2). Having a drink at this point helps. Factorise out Γ(1/2) in the denominator and you're left with something like dtdϕ(sinϕ.e^-tϕ)/sqrt(ϕ) inside a double integral. Getting rid of the t-integral (second drink optional) will get you ϕ^-½/(1+ϕ²) at which you substitute x'= sqrt(ϕ), et voilà.
I heard it slightly involved complex numbers, ran as fast as I can.
11:18 well you know what they say, "small mistakes makes rockets crash to planets, or a heap of math nerds really angry."
Definitely rockets😂
I tried to come up with a general formula for the power is n(any positive integer) instead of 2 using your method
But I stopped at the last step
Integrate tan^a(x) from 0 to π÷2
Is there an easy way to do this?
Try either the beta function or residues
If you're not comfortable with the beta function, watch my video on the michael penn integral (and share it too plz!)
@@maths_505 it took me a while to find a simple video that proof the relationship between the gamma and the Betta functions but I fainaly found it
The integral from 0 to ∞ of
e^(i(x^a))(when a is any positive number greater than 1) is equal to
((Factorial((-1/(2*n)))*Factorial((((-2*n)+1)/(2*n))))+((I*Factorial((((-n)-1)/(2*n))))*Factorial(((1-n)/(2*n)))))/((2*n)*Factorial((-1/n)))
Not the simplest way to write it but this is the best I could
I wrote it in terms of the factorial function instead of the gamma function because it Confuses me sometimes
It surprised me that if the power changed the values of the two integrals won't be the same any more
I am not sure if the formula work with complex numbers
When I substituted with I in the formula it gave me a finite value but when I computed the integral in Wolfram alpha to compare the answers it didn't give me anything
Again thank you for the video
Another way to do it, its by studying
I(t)=(integral from zero to infinity)exp(-tx^2)
the feynmann way. Then you substitute t=i and you evaluate the real and imaginary parts
Its way faster (but its less cool)
If you take a change of variable the equal the equal x √ small I
t=xi^.5
11:19 💀 i thought bro messed up something too important huh it was just a constant but really I was scared
Great solution!
I personally would have to liked it if you'd taken a little bit of time to appreciate the fact that the integral of the sqrt(tanx) is in fact equal to its reciprocal over the interval from 0 to π/2, something which has been astounding for me ever since I found it out! It's crazy!!!
I think I did that a while ago so I just pointed it out here as part of the phase shift
King’s property crying in the corner
Question: when you evaluate the limits, you say that when x->infinity, exp(-x*(t-i)) goes to zero. The outer integral w.r.t. t goes from 0 to infinity. The limit above does not converge when t=0. Is it ok to say the case where t=0 is excluded in the outer integral because you can always remove null-sets from integration domains and still get the same result, i.e. the integral over t>=0 and t>0 are the same?
You didn’t have to do all of that. Suppose that the parametrised gamma function integral is denoted as J. Which you call I_1 (i will call it J). The value of J(x) = sqrt(π) / sqrt(x)
Our original intgral I is just 1/2 ( J(i) )
We can just substitute x = i in the above formula and get the answer! There is no need to do a double integral other than to flaunt you integration skills of course!
interesting
I've seen you use this trick of finding the representation of a function in some different way, and maybe you know this but im surprised you haven't pointed it out, its just a Laplace transform. To illustrate, maybe you can do one integral you did a while back with this method, the int from 0 to inf of 1/(1+x^4) dx. i.e find a representation of 1/1+x^4 in terms of a certain laplace transforms and use it to get the answer, pretty neat stuff.
it took me a while to find a simple video that proof the relationship between the gamma and the Betta functions but I fainaly found it
The integral from 0 to ∞ of
e^(i(x^a))(when a is any positive number greater than 1) is equal to
((Factorial((-1/(2*n)))*Factorial((((-2*n)+1)/(2*n))))+((I*Factorial((((-n)-1)/(2*n))))*Factorial(((1-n)/(2*n)))))/((2*n)*Factorial((-1/n)))
Not the simplest way to write it but this is the best I could
I wrote it in terms of the factorial function instead of the gamma function because it Confuses me sometimes
It surprised me that if the power changed the values of the two integrals won't be the same any more
I am not sure if the formula work with complex numbers
When I substituted with I in the formula it gave me a finite value but when I computed the integral in Wolfram alpha to compare the answers it didn't give me anything
Again thank you for the video