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.
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!
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
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).
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.
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!
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.
For a better understanding of what is going on (other than just examples), I recommend this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-3LsXWPzlOhQ.html
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!!
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.
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.
+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
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.
@ 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?
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.
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!
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.
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.
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
