@@Chr15T a phase shift is when you use a substitution for a trig function pi/2-theta. This is done usually when a trig function is inside the natural logarithm. I knew it right away because the natural logarithm part has an initial substitution of 2^x-1= tan^2(u).
We can calculate it without trigonometric functions but we need two substitutions First substitution t^2 = 2^x-1 Second substitution u=1/t After first substitution we will get integral I_{1} After second substitution we will get integral I_{2} Now we can add them to get 2I = I_{1}+I_{2}
A very impresssive calculation ! The first substitution is far from being obvious. By the way : MATHEMATiCA ,which in general is very good for integrals , was not able to find the result.
I actually thought of that substitution right away, but I probably wouldn’t have thought of the rest of the tricks down the line like the cofunction force or equivalent addition of the two integrals
Well it's all kinda implicit when you integrate using a phase shift. You see the integration is still being carried out from zero to pi/2. Try working through your substitution again.
Bro how... how do you come up with these substitutions?! And who has the patience to think about one integral for that long? I would have tried u = √(2^x-1) or u = ln(2^x-1) or even u = ln(√(2^x-1)). But never freaking 2^x-1 = tan²(u) .....
I used substitution 2^x - 1 = u^2 then i divided interval of integral into [0,1] and [1,infinity] Int integral [1,infinity] i substituted u=1/w and added both integrals
Cool substitution. I'm not sure I could have come up with it. One note, your 'n's look way more like 'u's than anything else. Seeing tau and lu functions here 😂
Great substitution I knew right away when putting a trig function inside a logarithmic function that this might turn into a phase shift integral.
Please elaborate.
@@Chr15T elaborate on what?
@@moeberry8226 What is a phase shift integral and how did you know that right away?
@@Chr15T a phase shift is when you use a substitution for a trig function pi/2-theta. This is done usually when a trig function is inside the natural logarithm. I knew it right away because the natural logarithm part has an initial substitution of 2^x-1= tan^2(u).
We can calculate it without trigonometric functions but we need two substitutions
First substitution t^2 = 2^x-1
Second substitution u=1/t
After first substitution we will get integral I_{1}
After second substitution we will get integral I_{2}
Now we can add them to get
2I = I_{1}+I_{2}
Absolutely amazing integral techniques i've ever seen, thank you so much.
It's amazing to see such a great substitution, it require a lot much patience
A very impresssive calculation ! The first substitution is far from being obvious. By the way : MATHEMATiCA ,which in general is very good for integrals , was not able to find the result.
I actually thought of that substitution right away, but I probably wouldn’t have thought of the rest of the tricks down the line like the cofunction force or equivalent addition of the two integrals
How the heck did the author come up with this brilliant substitution? Practice definitely makes perfect.
What a crazy integral , love it 😍
I'll try this another way if i can
Mistake at 2:35: "tan x" should be "tan u"
Corrected at 4:33
You have to know these things 😊
Impressive!!
6:10 there's a minus missing before the integral after the Substitution.
No it's not missing. If you switch up the order of limits of integration, it cancels out that negative sign
@@maths_505 Yeah but you didnt switch the order of integration.
Well it's all kinda implicit when you integrate using a phase shift. You see the integration is still being carried out from zero to pi/2. Try working through your substitution again.
@@maths_505 oh okay i see. my bad.
Bro how... how do you come up with these substitutions?! And who has the patience to think about one integral for that long?
I would have tried u = √(2^x-1) or u = ln(2^x-1) or even u = ln(√(2^x-1)). But never freaking 2^x-1 = tan²(u) .....
There's never any harm in trying a substitution
Great video! I wanted to see you re-substitute u for x, but still fantastic!!!!
Are there other methods to solve this?
Yes ofcourse but this one's just the most awesome
I used substitution
2^x - 1 = u^2
then i divided interval of integral into [0,1] and [1,infinity]
Int integral [1,infinity] i substituted u=1/w
and added both integrals
Impressive
Cool substitution. I'm not sure I could have come up with it. One note, your 'n's look way more like 'u's than anything else. Seeing tau and lu functions here 😂
😂😂😂
ASTOUNDING!
nice trick
Very easy!
In my country, we used to solve this kind of integrals while at kindergarten!!
Just joking.
Awesome!! Thanks for sharing.
What about of convergence? You don`t proove that. Maybe, this integral does not converge, then we can`t substract and simplify