15:32 Both integrals along the vertical sections should have an addition factor of i because the parameterisation z= \pm R + iy results in the differential being dz = i dy, naturally. Obviously this doesn’t affect the cancellation but it is a necessary nitpick as my first thought when seeing the initial u =tanx substitution carried out was to do a semicircular contour with a branch cut on the negative imaginary axis, making sure to go around the origin with another small semicircle (of radius epsilon) traversed in a clockwise direction. In that situation, the positive and negative real axis contours would have different parameterisations and the differential would therefore be important to get right.
Alternatively, at 2:20 you can actually do the standard half-circle contour in the upper half plane as the radius R goes to infinity to get the correct result. After watching the video I went and evaluated that myself. I think it is simpler, but not as interesting as the additional substitution he does which shows a contour I haven't seen before.
@@alainleclerc233 Taking f(z) = z^i/(z^2 + 1) the denominator has only two poles, at i and -i. Only i would be in the enclosed region. Did you not click on the timestamp I linked?
No I did’n’t click on your timestamp because I thought you referred to the intégral from -infinity to infinity which was consistant with your half-circle proposal. Such a function icos(t)/(2cosh(t) has an infinity of poles in your contour
Wolfram Alpha gives; integrate (tan(x))^(i) from x=0 to x=pi/2 = π sinh(π/2) csch(π)≈0.62602 which is simply equiv. π /[e^( π/2)+e^(- π/2)] as Michael given at end.
4:32 I think, you should have mentioned that it works only if integral converges. For example, somebody may think that integration of sinx from -∞ to +∞ can be done in such manner due to symmetric limits and odd integrand (no, this integral diverges and is not equal to zero)
It still “doesn’t work” even if it converges; f(x)=x integrated from -inf to inf has a Cauchy principle value of zero, but it (the integral) doesn’t actually exist.
Alternately Beta Function identities make mince meat of this problem. Also you can expand it as a series but that requires some work to express it in terms of sechx
Agree ! It’s better only because it’s nicer. The intégration of the vertical segments is much easier to handle with the shorter notation as only half of the expressions disappear while the other half does not but has à zéro limit while R goes to infinity.
I wrote the numerator of the u integral as e^(i ln(u)), and used a keyhole contour with the ln branch cut on the positive real axis. The integral below the branch cut is -e^(-2 pi) times the integral (the original) above the cut. There are simple poles at +/- i.
Wolfram alpha calculates Laplace transform of cos(t)/cosh(t) with digamma function (derivative of logarithm of Gamma) If we plug in s=0 in Laplace transform of cos(t)/cosh(t) we will get the answer This digamma function has complex argument
You could have already solved it for general complex s as in I(s)=int_0^infinity dx x^s/(x^2+1) by considering the integral int_C z^s/(z^2+1) dz where C={-R+i0,R+i0} + {|z|=R,Im(z)>0} and R->infinity, choosing the cut of x^s to be (-infinity,0). The integral over the arc |z|=R vanishes in the limit R->infinity for Re(s) € (-1,1) and for those s also the integral over the semicircle of radius epsilon at z=0 vanishes as epsilon->0. The integral over (-infinity,0) is the same as the one over (0,infinity), except it picks up a phase factor of e^{i*Pi*s). On the other hand the residue is easily obtained to be e^{i*Pi/2*s}/(2i) for the pole at z=i. Hence (1+exp(i*Pi*s))*I(s)=2Pi*i*exp(i*Pi/2*s)/(2i)=Pi*exp(i*Pi/2*s) which gives I(s)=Pi/2*1/cos(Pi*s/2).
couldn't you also argue that for the thin sides of the contour, in the limit as R goes to inffinity, the e^R in the denominator basically kills the entire thing?
Why are ppl making those vertical integrals so complicated and handwavey? The denominator can be written as 2*cos(R+/-i*pi) and the whole integrands cancel out to i/2 (he missed a factor of i from the contour jacobian). The opposite bounds then eliminate the two.
Prof right at the very end you made a mistake of not including the +1 with cos(ipi) Because 1+cos(ipi) = 1 + (e^-pi+e^pi)/=(2+e^-pi+e^pi)/2 The final answer I got for that beautiful integral is pi(e^pi/2+e^-pi/2)/(2+e^pi+e^-pi)....
I think he meant "algebraic" as a shorthand for "non-trigonometric" rather than an actual algebraic expression. As for why he didn't just divide by 1+cos(i pi), he *did* do that.
@@xizar0rg The thing is I computed both results and they are different. The division of the residue and (1+cos(i*pi)) results in 0.68... while pi/(e^(pi/2)+e^(-pi/2)) is 0.626... I didn't do the simplification by hand, maybe he made a simple mistake, but I thought that maybe there is another calculation that I'm completely missing
@@thandolwethu3430 okay i will give it a try but please keep in mind im not a native english speaker and i have only studied math in school until 12th grade, im self taught. i hope it still helps^^ well about the first substitutions: the original integrand has an i in the exponent which makes this integral hard in the first place. so the substitution u=tanx at least made this easier by not having to deal with trigonometric functions anymore and also when i first watched this video i honestly thought he'd use the beta function from that point on or go straight into complex analysis/residue theorem. but the second substitution is brilliant because by letting that u=e^t we now have the i composed into the exponential function and that means we can use eulers formula to split it into the real and imaginary parts. now we have to only deal with 2 real integrands and a factor of i which doesnt really bother us since it can be pulled outside the integral. im not sure how to explain the paramatrizations but you might notice that all those 4 path integrals are actually being parametrized since we are integrating over a path where 1 variable is changing. lets take a look at "our original integral" which goes from -R to R. here the changing variable is R while the imaginary part is constant at i*0 as we integrate along that path, so we could parametrize it like this: z=t+i*0 and dz=dt where t is an element of the interval [-R;R] and viola we parametrized it so that we can now integrate from -R to R since our variable t is an element of that interval (it feels like a substitution) now lets look at another example, the path integral on the right side of the rectangle: here you can see that the R stays constant while the imaginary part varies from 0 to pi as we "go" along the path so our parametrization would be: z=R+i*t and dz = i*dt where t is an element of the interval [0; pi] now we can integrate from 0 to pi since our function has been parametrized with the that variable t
Is there a general method for deciding when to make the upper part of the contour a rectangle rather than a semi-circle? I usually see a semi-circle in these types of problems on TH-cam. Also, how are you inspired to use the e**t substitution?
Hey, Professor Penn. Would you please explain in full detail why the values of the integrals along the vertical parts of the contour are equal and opposite? Thank you.
Beware that as the radius of the semicircle goes to infinity, the contour will include infinitely many singularities. That means that in the residue theorem, the sum of residues becomes an infinite series, which is not easy to calculate. The rectangle contour only contains one singular for all values of R
You can use an upper half circle as the contour of integration if you don't do the second substitution and instead just look to integrate u^ i / (1+u^2) = e^(iln(u))/(1+u^2) from 0 to infinity. In this case the upper circle part goes to 0, and the radius from -R to 0 is just a multiple of the integral from 0 to R (which is the integral we want to solve for). The only reside is at i.
If you let x^i be multivalued for real x then eulers formula e^ix = cosx + isinx which was used in the video also doesn't hold anymore because you would get e^ix=e^2kpi * (cosx+isinx)
@@maxhofman6879 yes.why not. If k=0,,then we have Euler Formel. There is no restriction on k. Why not e^ix = e^2k pi(cos x +I sin x). If multi value is allowed, the result in this video may look like (…)e^(2pi k) for all integer k
at around 5 minutes you cancel out the odd function, my question is how do we know the -infinity and positive infinity are of the same magnitude? does it matter?
the -inf to inf interval is taken as the limit of symmetric intervals around zero. this means that at all values of R, the odd functions will cancel thus they do in the limit
@@yoav613 wait does it converge? I thought periodic trigonometric functions like sinx in the numerator didn't converge Maybe I'm thinking of something else Because like it would be just, like, snaking over and under around 0 wouldn't it? Oh but it will be getting close and closer to 0, right?
Hi. We are used to taking the big semicircle connecting -R to R as part of the contour. I guess it wouldn't work in this case and the integral over it would not be zero. What is the reason?
The semicircle contour would include infinitely many more poles because of the periodicity of the exponential. You'd have all the poles along the imaginary axis in the upper plane. It's much easier using this rectangle because there's only one pole to deal with.
@@brandonwillnecker8060 that's not the main issue though. The function on the semicircle does not tend to zero as R tends to infinity. That wasn't clear to me in the first glance
Your comment that an integral of an odd function over (-inf,inf) is zero is not necessarily true, you need to show it. For example, f(x)=x integrated from -inf to inf does not exist.
Cancellation of the integrals over the “thin sides” of the contour is elementary algebra. There is no convergence to show in this problem because the integral over the two “wide sides” do not vanish as R goes to infinity
15:32 Both integrals along the vertical sections should have an addition factor of i because the parameterisation z= \pm R + iy results in the differential being dz = i dy, naturally.
Obviously this doesn’t affect the cancellation but it is a necessary nitpick as my first thought when seeing the initial u =tanx substitution carried out was to do a semicircular contour with a branch cut on the negative imaginary axis, making sure to go around the origin with another small semicircle (of radius epsilon) traversed in a clockwise direction. In that situation, the positive and negative real axis contours would have different parameterisations and the differential would therefore be important to get right.
Correct !
Alternatively, at 2:20 you can actually do the standard half-circle contour in the upper half plane as the radius R goes to infinity to get the correct result. After watching the video I went and evaluated that myself. I think it is simpler, but not as interesting as the additional substitution he does which shows a contour I haven't seen before.
No, since the number of singularities in the half-disc increases to infinity with R while Michael’s rectangular contour contains only 1 singularity.
@@alainleclerc233 Taking f(z) = z^i/(z^2 + 1) the denominator has only two poles, at i and -i. Only i would be in the enclosed region. Did you not click on the timestamp I linked?
No I did’n’t click on your timestamp because I thought you referred to the intégral from -infinity to infinity which was consistant with your half-circle proposal. Such a function icos(t)/(2cosh(t) has an infinity of poles in your contour
Wolfram Alpha gives;
integrate (tan(x))^(i) from x=0 to x=pi/2
= π sinh(π/2) csch(π)≈0.62602
which is simply equiv. π /[e^( π/2)+e^(- π/2)] as Michael given at end.
4:32 I think, you should have mentioned that it works only if integral converges. For example, somebody may think that integration of sinx from -∞ to +∞ can be done in such manner due to symmetric limits and odd integrand (no, this integral diverges and is not equal to zero)
It still “doesn’t work” even if it converges; f(x)=x integrated from -inf to inf has a Cauchy principle value of zero, but it (the integral) doesn’t actually exist.
I was really fascinated. Every time I solve Improper integrals without Definite function in this way, I feel like I'm doing magic.
Alternately Beta Function identities make mince meat of this problem. Also you can expand it as a series but that requires some work to express it in terms of sechx
for a bit shorter notation we can use the hyperbolic cosine: (exp(x)+exp(-x))/2=cosh(x). it's easy to see that cos(ix)=cosh(x)
Agree ! It’s better only because it’s nicer. The intégration of the vertical segments is much easier to handle with the shorter notation as only half of the expressions disappear while the other half does not but has à zéro limit while R goes to infinity.
I understand all the steps and it’s really fascinating. I just don’t know how people get so good at recognizing what substitutions to make.
You can solve the integral very easily using the beta function and the euler reflection furmula
I wrote the numerator of the u integral as e^(i ln(u)), and used a keyhole contour with the ln branch cut on the positive real axis. The integral below the branch cut is -e^(-2 pi) times the integral (the original) above the cut. There are simple poles at +/- i.
the result can be expressed further as
pi/2 sech(pi/2)
Why does it say you need a full math major to understand. Teach us complex analysis, professor michael. I love you , and i will donate.
The final answer should be
pi*(e^pi/2+e^(-pi/2))/(2+e^(-pi)+e^pi).
I’m pretty sure the final answer is actually pi/(e^(pi/2)-e^(-pi/2))
@@JM-us3fr I confirm!
Ah, I needed a double take as well: your denominator is a perfect square: e^(-pi)+2+e^pi = (e^(-pi/2)+e^(pi/2))^2
19:19
A very nice and interesting example !
Wolfram alpha calculates Laplace transform of cos(t)/cosh(t) with digamma function (derivative of logarithm of Gamma)
If we plug in s=0 in Laplace transform of cos(t)/cosh(t) we will get the answer
This digamma function has complex argument
You could have already solved it for general complex s as in I(s)=int_0^infinity dx x^s/(x^2+1) by considering the integral int_C z^s/(z^2+1) dz where C={-R+i0,R+i0} + {|z|=R,Im(z)>0} and R->infinity, choosing the cut of x^s to be (-infinity,0). The integral over the arc |z|=R vanishes in the limit R->infinity for Re(s) € (-1,1) and for those s also the integral over the semicircle of radius epsilon at z=0 vanishes as epsilon->0. The integral over (-infinity,0) is the same as the one over (0,infinity), except it picks up a phase factor of e^{i*Pi*s). On the other hand the residue is easily obtained to be e^{i*Pi/2*s}/(2i) for the pole at z=i. Hence (1+exp(i*Pi*s))*I(s)=2Pi*i*exp(i*Pi/2*s)/(2i)=Pi*exp(i*Pi/2*s) which gives I(s)=Pi/2*1/cos(Pi*s/2).
couldn't you also argue that for the thin sides of the contour, in the limit as R goes to inffinity, the e^R in the denominator basically kills the entire thing?
No. Because the cos function for complex variables is also unbounded. So the numerator goes to infinity as well
@@MathwithMing in the interval under consideration it is bounded
Yes this also works
@@MathwithMing Here it works when you develop everything in trig and hyperbolic functions of real numbers
Why are ppl making those vertical integrals so complicated and handwavey? The denominator can be written as 2*cos(R+/-i*pi) and the whole integrands cancel out to i/2 (he missed a factor of i from the contour jacobian). The opposite bounds then eliminate the two.
Even if the last 2 integrals didn't cancel, they would still vanish as R->infinity, right?
yes. we can also use integral inequalities to arrive at 0.
Nice T-Shirt, Paul !!
Prof right at the very end you made a mistake of not including the +1 with cos(ipi) Because 1+cos(ipi) = 1 + (e^-pi+e^pi)/=(2+e^-pi+e^pi)/2
The final answer I got for that beautiful integral is pi(e^pi/2+e^-pi/2)/(2+e^pi+e^-pi)....
I factored it out.same answer
Teacher: Today I've got a nice integral.
The integral: 👁👄👁
I don't see the algebraic equation in the final step. Why can't you just divide the residue result with the (1+cos(i*pi)) after being expanded?
I think he meant "algebraic" as a shorthand for "non-trigonometric" rather than an actual algebraic expression. As for why he didn't just divide by 1+cos(i pi), he *did* do that.
@@xizar0rg The thing is I computed both results and they are different. The division of the residue and (1+cos(i*pi)) results in 0.68... while pi/(e^(pi/2)+e^(-pi/2)) is 0.626... I didn't do the simplification by hand, maybe he made a simple mistake, but I thought that maybe there is another calculation that I'm completely missing
That does need to happen.
@@krisbrandenberger544 what?
@@christianbuzzio1002 The division of (1+cos(i*pi)) did not happen when it needed to.
how do you come up with the subs? that's amazing - i am crushing on the math
you mean the first subs for tanx or the parametrizations?
@@xulq everything - all the subs and approach
@@thandolwethu3430
okay i will give it a try but please keep in mind im not a native english speaker and i have only studied math in school until 12th grade, im self taught. i hope it still helps^^
well about the first substitutions:
the original integrand has an i in the exponent which makes this integral hard in the first place. so the substitution u=tanx at least made this easier by not having to deal with trigonometric functions anymore and also when i first watched this video i honestly thought he'd use the beta function from that point on or go straight into complex analysis/residue theorem. but the second substitution is brilliant because by letting that u=e^t we now have the i composed into the exponential function and that means we can use eulers formula to split it into the real and imaginary parts. now we have to only deal with 2 real integrands and a factor of i which doesnt really bother us since it can be pulled outside the integral.
im not sure how to explain the paramatrizations but you might notice that all those 4 path integrals are actually being parametrized since we are integrating over a path where 1 variable is changing. lets take a look at "our original integral" which goes from -R to R. here the changing variable is R while the imaginary part is constant at i*0 as we integrate along that path, so we could parametrize it like this:
z=t+i*0 and dz=dt
where t is an element of the interval [-R;R]
and viola we parametrized it so that we can now integrate from -R to R since our variable t is an element of that interval
(it feels like a substitution)
now lets look at another example, the path integral on the right side of the rectangle:
here you can see that the R stays constant while the imaginary part varies from 0 to pi as we "go" along the path
so our parametrization would be:
z=R+i*t and dz = i*dt
where t is an element of the interval [0; pi]
now we can integrate from 0 to pi since our function has been parametrized with the that variable t
maybe someone else might explain it much better im sorry^^
@@xulq you explained well thank you
Is there a general method for deciding when to make the upper part of the contour a rectangle rather than a semi-circle? I usually see a semi-circle in these types of problems on TH-cam.
Also, how are you inspired to use the e**t substitution?
Kinda glad I did physics ;)
Does Wolfram's Mathematica get this right?
pi/2 / cosh(pi/2) is elegant
I think (pi/2) sech(pi/2) looks better but yeah, it's nice. It's also
(pi * e^(pi/2)) / (1 + e^π)
QNCUBED3 would do it much better!
Hey, Professor Penn. Would you please explain in full detail why the values of the integrals along the vertical parts of the contour are equal and opposite? Thank you.
Only the real parts are equal and opposite. The imaginary parts are not : they are equal but have a zero-limit when R goes to infinity
@@alainleclerc233 Thank you!
Could a semicircle be used as our contour?
Beware that as the radius of the semicircle goes to infinity, the contour will include infinitely many singularities. That means that in the residue theorem, the sum of residues becomes an infinite series, which is not easy to calculate. The rectangle contour only contains one singular for all values of R
@@MathwithMing Ah right, I forgot about the periodicity of the denominator's solutions. Thank you for pointing it out.
You can use an upper half circle as the contour of integration if you don't do the second substitution and instead just look to integrate u^ i / (1+u^2) = e^(iln(u))/(1+u^2) from 0 to infinity.
In this case the upper circle part goes to 0, and the radius from -R to 0 is just a multiple of the integral from 0 to R (which is the integral we want to solve for). The only reside is at i.
In my knowledge, complex power function is multivalved, e.x, 1^i = e^(-2 k pi), for all integers k. Why is the result here singlevalued?
x^i = e^ln(x)i and because x is a real number it's more natural to only consider the real natural logarithm, which was done here
@@maxhofman6879 number 1 is real. but 1^i = e^(-2 k pi). Why should other values than 0 be excluded?
If you let x^i be multivalued for real x then eulers formula e^ix = cosx + isinx which was used in the video also doesn't hold anymore because you would get e^ix=e^2kpi * (cosx+isinx)
@@maxhofman6879 yes.why not. If k=0,,then we have Euler Formel. There is no restriction on k. Why not e^ix = e^2k pi(cos x +I sin x). If multi value is allowed, the result in this video may look like (…)e^(2pi k) for all integer k
@@maxhofman6879 th-cam.com/video/VQhuL74qNQA/w-d-xo.html
th-cam.com/video/QDsC6n_PZig/w-d-xo.html
at around 5 minutes you cancel out the odd function, my question is how do we know the -infinity and positive infinity are of the same magnitude? does it matter?
I would also like to know
the -inf to inf interval is taken as the limit of symmetric intervals around zero. this means that at all values of R, the odd functions will cancel thus they do in the limit
The integral sinx/(e^x+e^(-x)) from 0 to inf converges ,so this integral from -inf to inf is 0 because the integrand is odd.
@@yoav613 wait does it converge? I thought periodic trigonometric functions like sinx in the numerator didn't converge
Maybe I'm thinking of something else
Because like it would be just, like, snaking over and under around 0 wouldn't it? Oh but it will be getting close and closer to 0, right?
@@randomname7918 yes. You can also use : sinx/(e^x+e^-x)
the plot of (tanx)^i scares me.
Feynman found it problematic to derive integration of tangent. For QM involves the complex plane and not only the real.
Hi. We are used to taking the big semicircle connecting -R to R as part of the contour. I guess it wouldn't work in this case and the integral over it would not be zero. What is the reason?
The semicircle contour would include infinitely many more poles because of the periodicity of the exponential. You'd have all the poles along the imaginary axis in the upper plane. It's much easier using this rectangle because there's only one pole to deal with.
@@brandonwillnecker8060 that's not the main issue though. The function on the semicircle does not tend to zero as R tends to infinity. That wasn't clear to me in the first glance
Your comment that an integral of an odd function over (-inf,inf) is zero is not necessarily true, you need to show it. For example, f(x)=x integrated from -inf to inf does not exist.
That was my thought as well. th-cam.com/video/0SP2b0nFpwI/w-d-xo.html
Mr Penn so you are a dune fan too? It's my favourite book! :D
Can someone demonstrate how the very last step is correct? Is there a typo?
Yes. The correct answer is half of the one shown on the board
No typo. Cancel my comment
@@alainleclerc233 i dont get your comments?
Others seems to doubt the final step too
I think it is incorrect in the way it is presented
asnwer=t isit
If x = arctan(u) then dx = du/(u^2 + 1) not dx = du / u^2 + 1.
I have been along until minute 18:29, then he made to many steps at once kicking me out ...
Playing a bit hard and fast with some convergences/cancellations, I think.
Cancellation of the integrals over the “thin sides” of the contour is elementary algebra. There is no convergence to show in this problem because the integral over the two “wide sides” do not vanish as R goes to infinity
Please credit the discoverer of this powerful technique:
Augustin Louis Cauchy (1789-1857)
just say Cauchy bud everyone knows who he is by saying his name
@@kostasbr51 yes and a French one hahha France>>>all
@@norajcarnaj9207 I disagree. For a great man like him you give his full name.
@@Swybryd-Nation nobody cares dude even in France when we use one of his theorems just Cauchy is required and that's all
Is it just me or is this video really quiet lol