Hey everyone! I've decided to start making videos again. Let me know if there's any topic you want me to cover in anything related to math, physics, or engineering.
there seems to be a relatively scarce amount of info on this topic so thank you for shedding light on it with some worked examples. this has been added to the tool box
Really solid video! The situation that makes this technique the most insane (yet helpful) is when you introduce a factor of "1", and by "1" I mean something like "e^(b(x-2))" evaluated at b=0. That's a crazy way to generate a term of "x-2" through sheer force of will.
The second one can be done by breaking up the integral from 0,1/n, and infty. First part isn't linear but goes to zero if you send n to infinity (just a Taylor series). The second bit is linear so distribute integral to both the terms, reindex and cancel what you can. This method works in more generality for any exponents. Like -ax and -bx gives an answer of ln(b/a).
Hey, perhaps I'm a little late to this video. Just wanted to say thank you very much for such a simple video about this topic. since other explanations I have found have been too complicated for me. Now I can take on this sort of stuff with (relative!) ease compared to before!
Just a small tip. Instead of testing for the constant of integration in the last step you simply can use the fundamental theorem of calculus you know f(x)=int o to x f'(t)dt And substitute the original value of b (2 in the first integral or 7 in the second) in the upper bound and solve that definite integral.
I like the fancy font versus white/chalk board scribble. I guess I'm slower than others (processing time of information) and would ask you to slow your syllabic cadence. I can follow this at a slower pace and it's new, interesting. You obviously know your stuff and I thank you for this post. I don't know when I'll ever need to integrate sin(x)/x where x is [0,inf) as a math hobbyist but this stuff is kinda fun. Thank you for you contribution to free education!
If the question comes( x^2 -1 )/logx...how would you know that which number to assume parameter...like how to know that we have to solve a question with this method?
You have to make an "educated guess". After a couple of these integrals you get a feeling for it. But you don't really "know" immediately what works and what doesn't. If nothing helps, you have to start trying until it works.
@ 2:16 the variable of integration is x I.e: DX then shouldn't the integrand be partially differentiated with respect to y since y is a parameter here?
I'm going to join the chorus of people asking about how this video was made (i.e. how was this animated). It is incredibly slick for a YT math video from ten years ago. This sort of thing isn't too hard these days with Manim, but this video predates Manim so I'm very curious about how it was done so seamlessly. Please give us some insight. Thanks!!
You can't integrate sin(x^2) by this method (or at least, I don't know how to), but there are other functions you can integrate that aren't necessarily quotients. You can integrate x^n cos x or x^n sin x for instance. A more complicated integral includes ln sin x from 0 to π/2, or more generally ln(a^2 - cos^2 x) from 0 to π/2.
So if I understand correctly, you can’t evaluate the integral of sinx/x (let’s say from 0 to 1) with this trick. How can I tell whether using this method will help me or not?
You can evaluate sin x/x using this. Not from 0 to 1, because that cannot be reduced to elementary functions, but from 0 to ∞. It's actually a famous application of this. I made a comment about it already, but here it is for your convenience: The second example can be used to calculate ∫sin x/x from -∞ to ∞ with a little tweaking. The latter is a famous application of DUIS, although the substitution usually made seems to be very counterintuitive to me. E.g: pg 3 of www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf shows how it is usually done. The substitution seems to come out of nowhere. But by writing sin with complex numbers, the second example in the video provides an intuitive way of calculating the famous integral. sin x = (e^ix - e^-ix)/2i, which is very similar to the numerator in the example's integrand. We can transform the example by replacing e^-x by e^ix. The 2i is inconsequential since it can be factored out of the integral. Thus we evaluate: I = ∫(e^ix - e^-bx)/x from 0 to ∞ Of course, I'(b) remains the same and so I'(b) = 1/b and I(b) = ln(b) + C. In other words, everything proceeds as in the example. To determine C, we set b = -i, which makes the integrand. Replacing in our equation, we have: ln(-i) + C = 0 -ln(i) + C = 0 C = ln i e^iπ = -1, famously, and i = √-1, so e^iπ/2 = i, and ln(i) = iπ/2. Thus C = iπ/2 The integrand = 2i sin x/x when b = i. Replacing in our equation: ln i + C = 2i∫sin x/x dx from 0 to ∞ This evaluates to iπ. Dividing by 2i to get ∫sin x/x gives you π/2. You can exploit the fact that both sin x and x are odd functions to show that ∫sin x/x from -∞ to 0 is the same as the same integral from 0 to ∞. This means that the integral from -∞ to ∞ is twice that from 0 to ∞, which is π.
quick question - if you integrate 0, the result can be C right? Since C differentiated would be 0... so when you're substituting b = 0, the integral doesn't necessarily equal to 0, it can still be C right?
I'm 10 months late, but better late than never, right? In case you still have this doubt, what you're saying would be correct for INDEFINITE integrals. You get an antiderivative, let's say F(x) + c. Now what would you do for a definite integral? You'd evaluate the antiderivative at the upper and lower bounds, let's say 'b' and 'a' respectively, and subtract. So you'd get [F(b) + c] - [F(a) + c], which simplifies to F(b) - F(a). No c! Hope that helped!
In the previous equation, ln(y) = b ln(x), y is a function of b. Now you differentiate both sides with respect to b; using the chain rule on the left side gives you (ln(y(b)))' = ln'(y(b)) y'(b) = 1/y(b) dy(b)/db
but I have a question, Forget this if it is a silly one.., but really I can't understand this.. Is 'b' a constant or a variable, if it is a variable (as you show in your video) then why we put 'b' for '7' because '7' is not a variable??
b is a variable, but we are trying to solve a function for a specific value of b. In DUTIS we treat the integral as a specific instance of a function. So for example, he defined I(b), and then proceeded to calculate I(7). It's a bit like having f(x) = x^2. x is a variable, but f(2) = 2^2 calculates x^2 when x = 2. It's the same idea with I(b), except that instead of having x^2 as a function we have an integral, and in this specific instance we are calculating the integral when b = 7.
The explanation above pretty much covers it. We're making the integral we seek a special case of a more general function - then integrate the general function - then evaluate that general function for the special case of our original problem.
The sinc function sin(x)/x appears a lot in physics - diffraction, Fourier transforms, quantum mechanics, etc - and can only be integrated by this technique. We define I(b) as INTGRL[ sin(x)Exp(-bx)/x ] and proceed as in the video. I was never taught this method way back when I was in school, and was blown away when I first saw it on blackpenredpen.
The second example can be used to calculate ∫sin x/x from -∞ to ∞ with a little tweaking. The latter is a famous application of DUIS, although the substitution usually made seems to be very counterintuitive to me. E.g: pg 3 of www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf shows how it is usually done. The substitution seems to come out of nowhere. But by writing sin with complex numbers, the second example in the video provides an intuitive way of calculating the famous integral. sin x = (e^ix - e^-ix)/2i, which is very similar to the numerator in the example's integrand. We can transform the example by replacing e^-x by e^ix. The 2i is inconsequential since it can be factored out of the integral. Thus we evaluate: I = ∫(e^ix - e^-bx)/x from 0 to ∞ Of course, I'(b) remains the same and so I'(b) = 1/b and I(b) = ln(b) + C. In other words, everything proceeds as in the example. To determine C, we set b = -i, which makes the integrand. Replacing in our equation, we have: ln(-i) + C = 0 -ln(i) + C = 0 C = ln i e^iπ = -1, famously, and i = √-1, so e^iπ/2 = i, and ln(i) = iπ/2. Thus C = iπ/2 The integrand = 2i sin x/x when b = i. Replacing in our equation: ln i + C = 2i∫sin x/x dx from 0 to ∞ This evaluates to iπ. Dividing by 2i to get ∫sin x/x gives you π/2. You can exploit the fact that both sin x and x are odd functions to show that ∫sin x/x from -∞ to 0 is the same as the same integral from 0 to ∞. This means that the integral from -∞ to ∞ is twice that from 0 to ∞, which is π.
+Mayur Gohil I think it's fine. Considering the fact that many mathematicians have used different notations for the same concept. Newton used fluxions to describes derivatives rather then Liebniz notation or Euler notation for their derivatives. I'ts all just preference just like using dummy variables. Though i do prefer the common notation for partial derivatives like yourself, i think the alternate notation beneath the improper integral is interesting and permissible. Though i do understand the fuss of confusion and the ambiguity you have, as it can sometimes be a hassle when working with different kinds of calculus textbooks and all may use different notation for the same concept.
Mayur Gohil notations don't matter if it has the same meaning I know calculus of variations blah blah but notations are notations need not be universal for an individual
I was dazzled by it too, first time I saw it. It gets clearer each time you watch someone do it. May I recommend th-cam.com/video/s1zhYD4x6mY/w-d-xo.html ?
Hey everyone! I've decided to start making videos again. Let me know if there's any topic you want me to cover in anything related to math, physics, or engineering.
Complex analysis and its use in integration?
can you make a video on Transistors?
yes you have come back brother we are waiting you .....good luck
Vectorial Calculus thanks!
Can you make videos on calculus from basics
Paige is perfect.... She's making me learn new complicated things........
a math video using LaTeX? I need more of these!!
Lots of math videos do this nowadays do this using Manim
there seems to be a relatively scarce amount of info on this topic so thank you for shedding light on it with some worked examples. this has been added to the tool box
This video helped my group tremendously for a graduate fluids problem. Thank you!
how so ? differential equation already had this trick as far as i know
I feel more powerful now
:>
This is.. gold!
use three dots
Yeah that is true
Really solid video! The situation that makes this technique the most insane (yet helpful) is when you introduce a factor of "1", and by "1" I mean something like "e^(b(x-2))" evaluated at b=0. That's a crazy way to generate a term of "x-2" through sheer force of will.
The second one can be done by breaking up the integral from 0,1/n, and infty.
First part isn't linear but goes to zero if you send n to infinity (just a Taylor series). The second bit is linear so distribute integral to both the terms, reindex and cancel what you can. This method works in more generality for any exponents. Like -ax and -bx gives an answer of ln(b/a).
A different set of tools :)
+Alejandro Pelcastre Feynman feelings :)
+Marcos Vinícius Petri Glad you know!
Reading his book now!
Ah, this is exactly why I looked it up. Feynman is my idol aha :D
Hey, perhaps I'm a little late to this video. Just wanted to say thank you very much for such a simple video about this topic. since other explanations I have found have been too complicated for me. Now I can take on this sort of stuff with (relative!) ease compared to before!
You can use \partial instead of \delta for partial derivatives.
Isnt it still just delta but cursive
That partial derivative sign tho
I think the name of the symbol is del
@@andrewolesen8773 it's dho
the hardest part of partial derivatives is the fricking sign^^
Yeah that pissed me off
your way of explaining is really great good job
Just a small tip. Instead of testing for the constant of integration in the last step you simply can use the fundamental theorem of calculus you know
f(x)=int o to x f'(t)dt
And substitute the original value of b (2 in the first integral or 7 in the second) in the upper bound and solve that definite integral.
6:00 why can we do that if en.wikipedia.org/wiki/Leibniz_integral_rule says we can use the method only if -inf < a , b < inf
I like the fancy font versus white/chalk board scribble. I guess I'm slower than others (processing time of information) and would ask you to slow your syllabic cadence. I can follow this at a slower pace and it's new, interesting. You obviously know your stuff and I thank you for this post. I don't know when I'll ever need to integrate sin(x)/x where x is [0,inf) as a math hobbyist but this stuff is kinda fun. Thank you for you contribution to free education!
You made a video explaining something no one else could
Wow. Thanks a bunch, super helpful video. Will have a think of topics I'd like you to cover.
Can you help me with video on vector spaces especially proving if a set is a space vector or not.
its pretty very simple to solve im hoping to over come with some mor examples..thank you
This is a mind blowing technique! Thanks!
isn't the integration and differantiation variable supposed to be different for the Leibniz rule to hold?
Nah, they'll just end up cancelling. To simply move the derivative into the integral, you need to make sure the integration bounds are constants.
If the question comes( x^2 -1 )/logx...how would you know that which number to assume parameter...like how to know that we have to solve a question with this method?
You have to make an "educated guess". After a couple of these integrals you get a feeling for it.
But you don't really "know" immediately what works and what doesn't.
If nothing helps, you have to start trying until it works.
@@ainzsama1565 hmmm
weirdly high quality for old video
You explained so clearly. Thank you Bro's :)
Wow!!! I never see those Ei and li before!!!
This helps! I am currently pursuing computer science and engineering...I got perfect at my calculus 2 test because of your video...thanks!
Thanks a lot dude, you cleared a great doubt of mine.
Shouldn't those deltas be "Del" s? (\delta vs \partial in latex)
Needed this for fourier transformations thank you
@ 2:16 the variable of integration is x I.e: DX then shouldn't the integrand be partially differentiated with respect to y since y is a parameter here?
I'm going to join the chorus of people asking about how this video was made (i.e. how was this animated). It is incredibly slick for a YT math video from ten years ago. This sort of thing isn't too hard these days with Manim, but this video predates Manim so I'm very curious about how it was done so seamlessly. Please give us some insight. Thanks!!
what a great video! Thanks guys much love
I now understand Feynman's Magical technique
this was magical
How'd you animate this? Its a great idea
It's*
1-1 is 0 quick maths
Wew this one was very good! I subscribed.
Does this trick only for a quotient? I noticed the examples have f(X)/g(X). Can you use this trick to integrate a function like sin(x^2)?
You can't integrate sin(x^2) by this method (or at least, I don't know how to), but there are other functions you can integrate that aren't necessarily quotients. You can integrate x^n cos x or x^n sin x for instance. A more complicated integral includes ln sin x from 0 to π/2, or more generally ln(a^2 - cos^2 x) from 0 to π/2.
THANKS bro,it is really helpful.
DUDE!!!.... You saved my ass!!.... Thank you so much!!!
So if I understand correctly, you can’t evaluate the integral of sinx/x (let’s say from 0 to 1) with this trick. How can I tell whether using this method will help me or not?
You can evaluate sin x/x using this. Not from 0 to 1, because that cannot be reduced to elementary functions, but from 0 to ∞. It's actually a famous application of this. I made a comment about it already, but here it is for your convenience:
The second example can be used to calculate ∫sin x/x from -∞ to ∞ with a little tweaking. The latter is a famous application of DUIS, although the substitution usually made seems to be very counterintuitive to me. E.g: pg 3 of www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf
shows how it is usually done. The substitution seems to come out of nowhere. But by writing sin with complex numbers, the second example in the video provides an intuitive way of calculating the famous integral.
sin x = (e^ix - e^-ix)/2i, which is very similar to the numerator in the example's integrand.
We can transform the example by replacing e^-x by e^ix. The 2i is inconsequential since it can be factored out of the integral. Thus we evaluate:
I = ∫(e^ix - e^-bx)/x from 0 to ∞
Of course, I'(b) remains the same and so I'(b) = 1/b and I(b) = ln(b) + C. In other words, everything proceeds as in the example. To determine C, we set b = -i, which makes the integrand. Replacing in our equation, we have:
ln(-i) + C = 0
-ln(i) + C = 0
C = ln i
e^iπ = -1, famously, and i = √-1, so e^iπ/2 = i, and ln(i) = iπ/2.
Thus C = iπ/2
The integrand = 2i sin x/x when b = i. Replacing in our equation:
ln i + C = 2i∫sin x/x dx from 0 to ∞
This evaluates to iπ.
Dividing by 2i to get ∫sin x/x gives you π/2.
You can exploit the fact that both sin x and x are odd functions to show that ∫sin x/x from -∞ to 0 is the same as the same integral from 0 to ∞. This means that the integral from -∞ to ∞ is twice that from 0 to ∞, which is π.
This is good stuff! Thanks!
The man is a superb mathematician who greatly simplified Leibnitz's Rule
What’s up with that partial derivative notation though?
quick question - if you integrate 0, the result can be C right? Since C differentiated would be 0... so when you're substituting b = 0, the integral doesn't necessarily equal to 0, it can still be C right?
I'm 10 months late, but better late than never, right?
In case you still have this doubt, what you're saying would be correct for INDEFINITE integrals. You get an antiderivative, let's say F(x) + c. Now what would you do for a definite integral? You'd evaluate the antiderivative at the upper and lower bounds, let's say 'b' and 'a' respectively, and subtract. So you'd get [F(b) + c] - [F(a) + c], which simplifies to F(b) - F(a). No c!
Hope that helped!
Does the rule of inserting the derivative in the integral apply without checking whether the integral coverges first ?
Thanks for making this - helped a lot
I'm only here because I watched Young Sheldon and heard this complicated things.
MAGICIAN SA same
Same
Hello brother.
Same here
Same
That is such a cool trick! thx
Really nice video. I do think that you are using an unusual way to express a partial derivative.
How did you get 1/y(dy/db) = ln(x) in the proof for dx^b/db = ln(x) x^b? (@3:15)
In the previous equation, ln(y) = b ln(x), y is a function of b. Now you differentiate both sides with respect to b; using the chain rule on the left side gives you (ln(y(b)))' = ln'(y(b)) y'(b) = 1/y(b) dy(b)/db
Felix Kunzmann Thank you Felix
Beautiful dude..
I shall add this technique to my collection Master Kenobi
Matthew Whitaker Is it possible to learn this power?
Hilbert Black Anything is available to be learn once you embrace the dark side of the integrals
This is so cool and clever!
Is there any clue that you should take the derivative under the integral?
Weeee!
Up goes my integral function
This is really late, but I just want to say this is a really nice video!
so the integral from 0 to 1 of (x^2-1)/ln(x)=ln3???
but I have a question, Forget this if it is a silly one.., but really I can't understand this..
Is 'b' a constant or a variable, if it is a variable (as you show in your video) then why we put 'b' for '7' because '7' is not a variable??
b is a variable, but we are trying to solve a function for a specific value of b.
In DUTIS we treat the integral as a specific instance of a function. So for example, he defined I(b), and then proceeded to calculate I(7). It's a bit like having f(x) = x^2. x is a variable, but f(2) = 2^2 calculates x^2 when x = 2. It's the same idea with I(b), except that instead of having x^2 as a function we have an integral, and in this specific instance we are calculating the integral when b = 7.
The explanation above pretty much covers it. We're making the integral we seek a special case of a more general function - then integrate the general function - then evaluate that general function for the special case of our original problem.
How 'bout a survey of line and surface integrals ... There's so many cases as it's confusing .... Thanks!
This is awesome, thank you!
Brilliant Explanations thanks
How can we integrate I'(b) = X^b w.r.t X .?
Anshuman Tripathy (x^b)/(ln(x))
great video
Very well done guys
Very cool
What are some examples of real world problems this solves? When would i ever see the function x^x ????
The sinc function sin(x)/x appears a lot in physics - diffraction, Fourier transforms, quantum mechanics, etc - and can only be integrated by this technique. We define I(b) as INTGRL[ sin(x)Exp(-bx)/x ] and proceed as in the video. I was never taught this method way back when I was in school, and was blown away when I first saw it on blackpenredpen.
The second example can be used to calculate ∫sin x/x from -∞ to ∞ with a little tweaking. The latter is a famous application of DUIS, although the substitution usually made seems to be very counterintuitive to me. E.g: pg 3 of www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf
shows how it is usually done. The substitution seems to come out of nowhere. But by writing sin with complex numbers, the second example in the video provides an intuitive way of calculating the famous integral.
sin x = (e^ix - e^-ix)/2i, which is very similar to the numerator in the example's integrand.
We can transform the example by replacing e^-x by e^ix. The 2i is inconsequential since it can be factored out of the integral. Thus we evaluate:
I = ∫(e^ix - e^-bx)/x from 0 to ∞
Of course, I'(b) remains the same and so I'(b) = 1/b and I(b) = ln(b) + C. In other words, everything proceeds as in the example. To determine C, we set b = -i, which makes the integrand. Replacing in our equation, we have:
ln(-i) + C = 0
-ln(i) + C = 0
C = ln i
e^iπ = -1, famously, and i = √-1, so e^iπ/2 = i, and ln(i) = iπ/2.
Thus C = iπ/2
The integrand = 2i sin x/x when b = i. Replacing in our equation:
ln i + C = 2i∫sin x/x dx from 0 to ∞
This evaluates to iπ.
Dividing by 2i to get ∫sin x/x gives you π/2.
You can exploit the fact that both sin x and x are odd functions to show that ∫sin x/x from -∞ to 0 is the same as the same integral from 0 to ∞. This means that the integral from -∞ to ∞ is twice that from 0 to ∞, which is π.
thanks.. very helpful
Excellent
So good
That's enough for me. I subscribe.
this is awesome and soo cool
nice tutorial. keep it up
demonstration the Differentiation under the Integral Sign Tutorial
does anyone have some exercises about this ? :p
check out blackpenredpen and flammable maths
I have a BSc in Physics and I didn't even know how to do this
same here, it was new to me too. I love it!
More videos like this, please.
Good
amazing! 11.
Yay! New integration tools! :))))
thank you
But you shouldn't derivate only x^b but (x^b) /ln(x)
thank you for help me.
Who made this video? Make more!
This..... I like this....
I get the feel tho, but who the hell would say ugly integral in that serious of a tone😂
Thanks
Your partial derivative notation is in appropriate.
but you have to agree the delta sign looks similar to del
But it does not stand to standards...
+Mayur Gohil I think it's fine. Considering the fact that many mathematicians have used different notations for the same concept. Newton used fluxions to describes derivatives rather then Liebniz notation or Euler notation for their derivatives. I'ts all just preference just like using dummy variables. Though i do prefer the common notation for partial derivatives like yourself, i think the alternate notation beneath the improper integral is interesting and permissible. Though i do understand the fuss of confusion and the ambiguity you have, as it can sometimes be a hassle when working with different kinds of calculus textbooks and all may use different notation for the same concept.
Actually, coincidentally my teacher for differential equations uses the delta for partial derivative
Mayur Gohil notations don't matter if it has the same meaning I know calculus of variations blah blah but notations are notations need not be universal for an individual
thanks
@ 2:15 I think you're differentiating w.r. to y
I came here just cause i watched young sheldon
Nice presentation, but you should have used a more simple multivariable function. I was confused with you rushing through x^b explanation
なるほど
Papa flammy thumbnail
For a better understanding of what is going on (other than just examples), I recommend this video: th-cam.com/video/3LsXWPzlOhQ/w-d-xo.html
I find this a little difficult
I was dazzled by it too, first time I saw it. It gets clearer each time you watch someone do it. May I recommend th-cam.com/video/s1zhYD4x6mY/w-d-xo.html ?
WHOA
I subscribed