integral of sin(x)/x from 0 to inf by Feynman's Technique
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- Опубліковано 21 вер 2024
- The integral of sin(x)/x from 0 to inf by using Feynman's technique (aka differentiation under the integral sign). This integral is also called the Dirichlet integral. Check out another example of Feynman's technique of integration: • Feynman's Technique of...
Zachary's page: philosophicalm... ,
integral of sin(x)*e^(-bx), • The appetizer, integra...
Another example, Integral of ln(x^2+1)/(x+1) from 0 to 1 by Mu Prime Math, • It took me 3 hours to ...
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This is so famous, i still remember 8 years ago, when my uni professor told me, there is psychiatric hospital for those who still try to find a primitive of sin(x) / x... lol
Kkkkk
Lol
For some reason , “lol” looks like mod(0)
@@camkiranratna do you mean absolute value?
@@deltaspace0 Absolute value is also called mod in some places.
You don't often see a man doing partial derivatives while wearing a partial derivative t-shirt.
hahahahaha! honestly, that wasn't planned.
blackpenredpen I just realised after reading this...
me 2 lollll
It seemed to me like some sort of band sign like Nike at first
what’s the difference between partial derivative and normal derivative?
I recall doing this integral many years ago. Back then we used contour integration. We chose the contour to be a semi-circle of radius R centered at the origin . The origin was indented and cotoured with a semi-circle of radius r. The semi-circle was located in the upper-half of the Cartesian plane. Complex integration in one of the most potent methods for dealing with such problems.
I agree, I solved this too in my first course of Applied Mathematics in college where we used complex analysis techniques
ua-cam.com/video/Ff4LRlflib0/v-deo.html
Yeah that's true, that's how I learnt it/saw it first
Integrales cerradas en variable compleja?
You can do integrals on complex bounds (lower/upper) 😮? Or is it Real bounds but integrated on Complex functions?
@@louisrobitaille5810 complex functions and complex bounds. Turns out that the path you take *mostly* doesn't matter!
When you sleep in class 14:01
More like when you blink in class :)
but the answer was spoiled in that part :D
when you struggle not to sleep
Idk why was the video cut? lol
Ahnaf Abdullah I wanted to add that explanation why b has to be nonnegative
I really enjoy watching you integrate! Relaxing and fascinating at the same time.
Isn't it!
PompeyDB it is!
it is, is not?
It's!
I really enjoy watching you disintegrate! Relaxing and fascinating at the same time.
Isn't it!
@@rehmmyteon5016 lmao
you told us not to trust wolfram and now you confirm your answer in wolfram. what am i supposed to do with my life now?
Dokuta Viktor trust no one
Dokuta Viktor Ask wolfram.
Dokuta Viktor only if it gives the same answer as what we got.
blackpenredpen what if what you got is by looking at Wolfram????
then don't get things from Wolfram but just check your answer with it.
The part where the constant C is determined by checking the limit of the function at infinity is very elegant. Beautiful proof. Of course, there are a lot of technical details that mathematicians would think about (is it correct to derivate inside the integral, exchange limit and integral, etc.). But this video is a great summary of the overall strategy. Very nice work!
.l
One of the best math videos I´v ever seen. Changing the function from x to b was a masterpiece.
Yes, Feynman was a brilliant mind
I see you have finally decided to clothe like a true mathematician, seeing your t-shirt involves partial derivatives. 👌
MeowGrump lolllll this is a good one!!!
asics = "Anime sane in corpore sano,"
"Sound mind/spirit in a sound body."
Anima not anime (but that's somehow relevant :))))
👌 looks like the partial derivative sign XD
So, it is soul eater then?
Hi, I just learned this technique over the summer. I was amazed. I used it to solve a problem from American Mathematical Monthly. It was fun, not only sending in a solution, but learning this amazing technique used by Feynman!
"And once again, pi pops out of nowhere!"
Wow. At the begining the integral with the exponential function looks more complicated, but that function allows to have a closed form and the Leibniz theorem is fundamental. Great work!
Lies again? So fat
18:12
I like the idea that, after going through all that, we figure out that the integral from 0 to infinity of sin(x)/x dx is equal to...
Some unknown value.
its not that unexpected though if you look at the function... its just looks very convergent.. (this can ofc be very deceiving)
@@antonquirgst2812 But there's the fact that as x grows larger, it tends to 0 because sin's at most 1 or -1.
@@createyourownfuture5410 yup - totally agree - x grows linear while sin(x) is periodic!
@@antonquirgst2812 Aaaand it approaches 0 from both sides
@@createyourownfuture5410 It actually approaches 1 from 0
him: "And now let's draw the continuation arrow with also looks like the integration symbol. That's so cool."
Me: "Ha."
I happen to remember just enough calculus to follow along. Interesting. Thank you.
Sean Clough yay! I am happy to hear!
the world needs more of this....
He's becoming self aware
How so? What did you notice?
isn't it?
"This is so much fun, isn't it?"
Sure.
lol
Feynman was a mathematician, physician and philosopher... super geniuce
Physicist*
@@juanpiedrahita-garcia5138 lol
genius , even as a joke it hurts my eyes
You always manage to make me click to watch you do integrals I've already done long ago!, but this integral of sinc(x) was really gorgeous. It's kinda the method for obtaining the the moments of x with the gaußian. I hope to see more of this kind.
Thanks for reminding me what I enjoyed about maths! It really is good fun to play around with calculus like this.
Wow, that was an heavy load! I never saw anything like that before...it'll take me a few days to digest the technique. Well done!
That's what she said
i am 17 years old and i am from india .............i am able to understand it clearly ......thank you sir , love you and your love for mathematics 😊
💀I’m in 11th starting trying to learn this as my physics part needs it.
How the heck do 2 people that didn't know eachother ' invent' calculus at the same time.Simply fascinating. This was awesome to watch, I now have a better understanding of how partial derivatives work. I now must go back and study calc shui I can come back and fully digest this.
Great video using Feynman's technique but would never tackle this integral in this way. Once you've applied the Laplace transform it's much easier to use Euler's formula and substitute sin(x) with Im (e^ix). Haven't read all of the comments but I'm sure this has already been mentioned
I'm familiar with using the Fourier transform to find the integral, but I don't quite see how you'd use the Laplace transform.
@@Sugarman96 - the Laplace transform is what the above video uses when creating his function I (b)
@@Sugarman96
I'(b) is the same negative laplace transform of sin(x) which you can use to easily find I'(b) instead of doing whatever he did.
I can't believe I just spent 20 minutes watching a video about integration and loving every second of it. A few years ago, I used to despise Maths
Using laplace transform and fubini's theorem this integral reduces to a simple trig substitution problem.
For the ones that want to dive into the details, I think we have to justify that the differential equation is defined for b in (R+*) in order for e^(-bx) to actually tend towards 0, then use the continuity of parameter integrals so that I(b) -> I(0) when b->0. Finally, the dominated convergence theorem gives us that I(b) -> 0 when b->inf. We conclude with the fact that arctan + pi/2 -> pi/2 when b->0, and uniqueness of the limit : both limits I(0) and pi/2 are equal ♡
Why would anyone think to add e^x thiugh this COMES OUT OF NOWHERE..what I thought to do was replace sinex with e^ix from Eulers formula..isn't thst smarter and more intuitive? I think he needs to justify where e^x cones from if anything it should be ln x he is adding nkt e^× since 1/× is the derivative of ln x not e^×..
Believe in Math; Believe in the Pens; Believe in Black and Red Pens.
yay!!!!
Yes, i did It and i got 10 in my integral calculus exam :') two months ago !
@@MrAssassins117 now 3 years ago lol
@@pranav2119 now 3 yrs and 14 hrs ago.
I still remember my first year in college. It was filled with so many wonderful moments. This was not one of them.
Best math teacher ever !!!
Now it's time for the Gamma function and some other Euler integrals ;>
Please do some putnam integrals
They are really tricky and also few tough integrals like these.
I love watching your integration videos.
This problem can be simply solved using complex integral (getting the answer directly without a piece of paper). However, I’ve to admit that the method introduced here is VERY SMART. Thank you!
This integral's easy. Just pretend that all angles are small, replace sin(x) = x, the x's cancel so you're left with the integral of 1 :D
Hard to justify with those zero to infy limits. ;-)
so, pi/2 \approx inf?
How can you pretend all angles are small? The angle goes to infinity o_O
@kikones34 : Yeah, that's the joke (note the ":D" grin at the end.). But it _does_ work for the _variable_-bound integral
int_{0...x} sin(t)/t dt
which, by the way, defines the standard mathematical function Si(x), the "sine integral" function, because you can then consider when all angles in the integration are small. If you take sin(t) ~ t then you say for _small_ x that
int_{0...x} sin(t)/t dt ~ int_{0...x} t/t dt = int_{0...x} dt = x
so Si(x) ~ x when x is small. And a Taylor expansion will show you that that makes sense, too:
Si(x) = x - x^3/(3.3!) + x^5/(5.5!) - x^7/(7.7!) + x^9/(9.9!) - x^11/(11.11!) + ...
so the first (lowest-order) term is x, thus at small x, Si(x) = x + O(x^3), meaning the rest vanishes like x^3.
@mike4ty4 Oh, sorry, I totally didn't get you were joking. I've been on a UA-cam trip of flat earther videos before watching this, so I was in a mindset in which I assumed nonsensical statements are actually serious and not jokes xD.. D:
"so lets draw the continuation arrow, which looks like an integral sign, that is so cool"... friends, this guy is pure gold !!!!
At 14:18 : You say that since e^-bx matters, the integral converges for all values of b >= 0. Well it's true for b > 0. The reasoning cannot work for b = 0 because it's slightly more complicated than that (but it converges too).
Counter example : Integral from 0 to infinity of e^-bx/x dx doesn't converge for b = 0.
Hi, this is Zachary Lee.
You are absolutely right to be concerned about the convergence at b=0. What you want to do is let b approach 0 from the right. If you want a rigorous explanation, check out Appendix A, on page 21 of this document:
www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf
Footskills here's the man!!!
Yeah that was a brief explanation haahahhahaha
And here I am again!!! Btw, great explanation!
PackSciences your counter example should be rearranged as (e^(-b x)-1)/x
Btw e^(-b x)/x diverges for all values of "b"
That was the most peaceful boss music I've ever heard. And it's definitely boss music when you're trying to integrate sin(x)/x
I knew this, but it is still awesome. More stuff like this pls!
We used to think that it is such a basic calculus skill for all college students, now it becomes a show and privilege. I hope it will bring more interests among the young generations.
These are so addicting to watch and I don't know why
His enthusiasm is contagious wish he was my calc professor back in the day I would have loved that class
Correct me if I'm wrong, I'm a bit rusty, but don't you need to prove uniform convergence before bringing the differentiation sign inside the integral?
I think you're right. I was thinking the same
Does it help? I am not an expert in the field (yet):
en.wikipedia.org/wiki/Leibniz_integral_rule
I found another way to solve his problem that feels more unique, alhough your solutions is much more straightfoward and intuative.
I started by doing everything the same up until you get to I'(t) = -integral of sintheta times e^(-t*theta)d theta. Afterward, I turned sintheta into Im(e^(i*theta)). Hrn I used exponent laws to combine the exponentials and and take the integral from 0 to inf. Then I took i tegral on both sides and evaluated I(inf) to get c=0. Then I evaluted I(0) = -Im(ln(0-i)) = pi/2.
Beautiful proof, thank you.
bruno edwards Yup, leibniz rule is very powerful.
I started to get interested in mathematics after seeing this integral before!
Thank you for giving me the solution :)
Can you recommend a good proof of Liebniz Rule to follow?
It seems like one of those simple/obvious things that would actually have an interesting/ instructive proof.
Instead use Fourier transform method, inverse Fourier transform of sampling function is gating function with parameters A and T
What's also very Interesting, we could also use *Lobachevsky's integral formula* :
*integral from 0 to +∞ of [ f(x) * (sin(x) / x) ] = integral from 0 to (π/2) of [ f(x) ]*
So our example:
integral from 0 to +∞ of [ (sin(x) / x) ]
has *f(x)=1* :)
Now we use Lobachevsky's integral formula:
*integral from 0 to +∞ of [ f(x) * (sin(x) / x) ] = integral from 0 to (π/2) of [ f(x) ]*
integral from 0 to +∞ of [ 1 * (sin(x) / x) ] = integral from 0 to (π/2) of [ 1 ]
integral from 0 to +∞ of [ (sin(x) / x) ] = integral from 0 to (π/2) of [ 1 ] = x | computed from 0 to (π/2) = (π/2) - 0 = (π/2)
*Answer:*
integral from 0 to +∞ of [ (sin(x) / x) ] = *(π/2)*
Mr Michael Penn made a video (entitled )
where he calculates that example using Lobachevsky's integral formula:
ua-cam.com/video/m0o6pAeCcJs/v-deo.html
"Lobachevsky's integral formula and a nice application."
Michael Penn
There's a simpler way of calculating this integral. This funcion is really famous, is the sinc function, and is the fourier representation of an ideal low-pass filter, a rectangular function.
The integration property of the Fourier transform tell us that the integral from minus infinity to infinity of a function in the time domain is equal to the frequency domain (or Fourier domain) representation of the function evaluated in 0.
So, to calculate this integral, you just calculate the Fourier transform and just evaluate in 0, which gives you Pi. Of course, because of the integration limits, you get the result divided by 2.
this result is used to prove the convergent of fourier series though
……人又相信 一世一生這膚淺對白
來吧送給你 要幾百萬人流淚過的歌
如從未聽過 誓言如幸福摩天輪
才令我因你 要呼天叫地愛愛愛愛那麼多……
If you know you'll know
Of course I know 😆
ahhh, that's why the intro song is so familiar, k歌之王 by Eason Chan!
There seem to be a few questionable parts about the video.
The most important one is at 20:41, it's understandable that as b goes to infinite and x does too, the value is 0, but if x=0 and b is infinity, then you have e^(0*infinity) which is questionable what it equals?
Another thing is that how can you assume that b=0 "works the same way" as positive b values, when negative b values don't give a convergent value, and mess up the whole thing? How can you say that 0 which is "on the border" works like positive values. But this probably can be answered easily, but the e^infinity*0 seems to me like no
The best ever demonstration i've seen.
I always thought this integral to be done by an algorithm based on the sum of trapezium areas which gives approximatively the same result as pi/2.
Really amazing demo.
The next question would be what is the primary function of integral of
sin(x)/x dx ?
This is the Dirichlet function and the Feynman technique is great way to solve it. Downside of Feynman technique is you cant plug and chug. The formulas have to be checked along the way for validity . Such is life. Thank you Pen(Black + Red)
Great videos, planning to recommend to my students but not a fan of notation x=inf or of plugging in x=inf. Students will do this without the understanding you have and will lead to some issues in calculating limits such as inf/inf =1. Please remember you're a role model :)
Deanna Baxter I always just plugged in infinity. Didn't lead to any misunderstandings. It's more cumbersome to take the limit, though it's technically correct. You first introduce indeterminate forms in order to avoid issues.
Rudboy
I agree, sadly sometimes students won't be lucky enough to get a grader who will be forgiving.
I one time did that and the grader goes
"While your final answer is correct, you can't just set something as infinity"
There was another part of the problem where I got the answer correct, and they go "your answer in this part is correct *AND* your math is right, but you weren't supposed to get it that way"
I ended up getting only half credit for that problem
This was an assignment where we had to do ten problems but only *two* of them would be selected at random and graded so one quarter of my grade on that went out the window
Needless to say, I was salty
Deanna Baxter if the students are interested in this integral in the first place, they should be ok and understanding this shorthand notation. Btw, a MIT professor also does that in his calc lectures for improper integral.
Here ua-cam.com/video/KhwQKE_tld0/v-deo.html
Thanks for the comment and thanks for watching!! :)
I did not know that Feynman won the Nobel prize for solving this integral, I thought it was for his contribution to QED
make video on the squeze theorem, I bet you can make it interesting and to show all techniques
Paul Montes dr. Peyam is actually going to do that soon
We learned the Feynman-Spell in Theoretical Physics 1 and Mathematical Methods of Physics (TU Berlin).
The teachers didn't mention, that this kind of integration and computation is the Feynman-Spell.
They called it Integration with respect to a Parameter b !
Great video. The e^(-bx) looks random until you realize that lots of these problems use the same parameterization.
The answer is actually 42 though. Proof: summing the positive and negative regions under the curve we get a conditionally convergent series. Add positive terms until you exceed 42, then add negative terms until you go below 42, then add more positive terms until you exceed 42 again, etc. The sum will converge to 42 so this is the value of the integral. QED.
Fantastic video! I was thinking literally just the other day that I hope you'd make a Feynman technique video and, as through magic, here it is! Would really love to see more videos about alternative / advanced techniques.
Your claim that the expression inside the integral is going to 0 when x approcheing to infinity is very problematic when you understand that we considering the case when b=0. Then, the integral wouldn't be convergent, so how can you explain that?
x approaches 0 from the right. With a weapon. Also discussed later in the comments
It's the first time I see this way of integration and I'm amazed!
Does theta stand for anything particular in Greek, relating to angles? Or is it just an arbitrary letter that has historically been used for representing angles similar to how x and y represent Cartesian coordinate variables?
Probably, the reason x/y/z are used for representing Cartesian coordinate variables, is that it is the trio of neighboring letters in the alphabet, that is LEAST likely to stand for anything in particular, and therefore they are letters used as wildcards.
Great video. Least I can do is thank you for a great explanation!
Thank you!!!
All the computations are only valid for b>0, because you need the exponencial to derive inside the integral under Lebesgue's domination Theorem. But at the end you do b=0. One further step is needed to show that I is continuous at 0. Note that this os not easy because |sin(x)/x| is not integrable, and therefore you cannot use standard continuity theorems as they require a domination hypothesis.
Hello Alejo. Yes, I agree, but that is exacly my point. You need a more demanding theory (such as Denjoy integrability, among other possibilities) to justify the calculus presented in the video.
I came in just because i saw the name Feynman
me too
You got it ,me too
You really save life via UA-cam
The content on your page is always so informative, and your excitement for the math you show is contagious. By the way, have you considered making a Patreon page? I would gladly support!
Also, how sneaky of you to wear the "Basic" shirt that has the lowercase-delta on it, foreshadowing the partial derivatives you use in the video.
LOL! Thanks!
In fact, that wasn't planned. lolllll
Thats not a lowercase Delta
Here comes the paid publishing
This technique is called "differentiating under the integral sign" and Feynman learned it from a book entitled Calculus For the Practical Man when he was a teen. Feynman didn't invent it but made it famous through his anecdotes.
The kids' laugh made me forget the stress of trying to understanding how you solve it. 😂😂😂😂😂😂😂😂😂
Mazed Suhaimi yay!!!!
small question, why cant the integral of -1/1 + b^2 be arccot(b) + c
ok nvm some things change but i end up getting pi/2 answer anyway so ig it can be it
Thank you for showing the trick with the e-function. Would not have seen this and could be very useful. When I did this problem for -inf to inf I did it with Fourier transformation by writing sinx/x as the fourier transformation of the rectangle function. After changing order of integration you get a delta distribution and the other integral collapses as well. Of course you get Pi at the end.
To be fair, Zach showed me (as I mentioned in the video).
Can you like a video twice? Just watched this again, and still awesome. Thanks for this!
The cut at 14:02 is kind of confusing.
thomas g Just got to it, I reckon he's solved it already, then started talking about his steps and realised it'd fit better with the part where he was previously (in his timeline) talking about it.
It's so he can outline that b has to be positive, and it probably makes the most sense to put the clip here
When u sleep on class
it spoiled the rest of the video
I’m convinced my teacher watches you. Whenever he says “isn’t it?” I think of BPRP
Who else reads his shirt as "partial asics"?
So at 20:13 you said something along the lines of: _something_/infinity = 0, then infinity * that 0 = 0. But what if you look at it like infinity * _something_/infinity, which would cancel out and yield _something_?
Lmao the "isn't it" in the thumbnail
Easier than solving for C is to write I integral from 0 to b of I' = I(b) - I(0)
Left-hand side is ok even if you use a different antiderivative, as long as the choice on the left is self-consistent. Then you can take limit b to infty and solve for I(0)
Love Feynman and this trick was cool and useful.
You now have another subscriber :)
Wonderful!!!!!!!!
Can you do the Gaussian integral -integral of e^-(x^2) from -inf to +inf
My mind was blown infinitely away
伊藤那由多 loll
In fact, the integral from minus infinity to infinity of sin(x)/x IS equal to Pi. It is called Dirichlet integral. Thanks, ChatGPT
20:35
Dont you get, if you integrate 0, another constant? Because the derivative of an Constant is 0 too
elp 486
It’s a definite integral of 0 from a to b, so there’s no area. : )
14:30
How can b be >= 0, shouldn't it be just > 0? If it is 0, then the cosine in the solved integral will not converge.
I'm late, but I'll try to explain (sorry I'm french). Here you're using a theorem (with hypothesis he's isn't precising) to be able to differentiate in the integral. This theorem also tells u thats I and I' are continuous on there interval of definition. This means that ye, b>0 for I' but when you get the expression of I, since I is continuous on [0,+inf[, u can say that I(0)=lim I(b) when b->0 and thats it
@@antred9157 Yes, b must be positive for the calculation.
And one still needs to show that the function I is continuous at 0. This is in fact not easy to do, since the absolute value |sin(x)/x| is not integrable on [0,+inf]. One usually proceeds as follows: Instead of integrating over [0,+inf], define I_n(b), for positive b, to be the integral over [0,n]. Now use Feynman's trick. Proceed as in the video, and finally show that I_n(n) goes to 0 as n goes to infinity to complete the proof.
K 歌之王?
Isaac TNT yes
blackpenredpen man of culture
sin(x)/x? More like "Super derivations that are always the best!" I know a lot of other comments say it, but I think this technique is just so cool, and it can take things beyond a lot of other integration videos. Thanks for sharing!
I found it easier to first complexify the integral and then use the Feynman Trick. Define F(z)=int from 0 to inf of e^-zx/x, so you have to find Im[F(-i)]. When differentiating and then integrating with respect to z you get F(z)=-ln(z)+C for Re[z]>=0, or F(-i)=ln(i)+C. One would usually try to calculate C by evaluating at 1, but it's easier to notice that for any positive real number x F(x) is an integral of a real function, and is therefore real, and ln(x) is also real, so C must be real too. This way when you take the imaginary part of both sides (which one has to do anyway), you get rid of C, killing two birds in one stone, so Im[F(-i)]=Im[ln(i)]=Im[iπ/2]=π/2
Where do I learn this power?
I don't get the 13:07 part, if b = 0, x doesn't matter anymore, coz bx is always 0, which makes e^(-bx) = 1, right? And I(0) is exactly what we're trying to solve.
Yes
Please do more video about the Feynman Techniques
Thanks a lot
Jack Chai ok
I'm curious about the very last step. Evaluating atan(0) should yield multiple solutions. How do you reconcile this? What justification do you have for selecting only the principal value?
The defined range of arctan(x) is arctan (x)∈[-π/2, π/2]. So therefore there is only one value!
Hi professor,
You are doing great...
Thank you!
In fact, 1/x=\int_{0}^{+\infty}{e^{-xy}dy}. We can change one dimensional integral \int_{0}^{+\infty}{sin(x)/xdx} to a two dimensional integral and exchange integral order. First, integral with respective with x, int_{0}^{+\infty}{1/(1+y^2)dy}=pi/2, this is the answer.
just noticed that his shirt is the partial derivative of asics
You are better than my professors!
When he reversed derivative on I(b) by integrating (14:45 min ) and evaluated result as b went to infinity and got zero for that limit-his argument failed. You only get zero if b>0, not if b=0. If b=0 you don't get zero as x goes to infinity-you get divergence
You can recover it, however, with just a little more rigor. Instead of evaluating I(0), find limb->0 I(b). Then, just keep using the limit notation until the end. The original integral is actually equal to limb->0 I(b) = Pi/2, so no real harm done.
why -e^bx * (cosx +bsinx) / (b^2 + 1) = 0? (if x goes to infinity) what if b = 0?
Yay i was waiting for this!
Notice that Sinx/X is defined on the whole IR line since Sinx is an odd function and by using the Taylor expansion: Sinx/x = 1-x²/3!+ x⁴/5!- x⁶/7! +.... Which is defined at X=0 and is equal to 1.
Isn't I'(b) undefined for b=0? It confuses me how you can make deductions about I(b) at b=0 from its differential when its differential is undefined at that point. Forgive me if this may sound dumb, the furthest I've been taught in school so far is integrating polynomials, but is there a way to justify this in a more rigorous sense or is it actually fine and I'm nit-picking over something irrelevant?
He is using the limit as x -> 0 of sin(x)/x which equals 1.
the function is undefined, but the derivative isn't
I(b) is defined in 0 but I'(b) isn't defined in 0
For example the function f(x)=sqrt(x) is defined in 0 but its derivate is not defined in 0 because f'(x)= 1/(2*sqrt(x))
more integrals using differentiation under the integral sign please!