4:21 I feel like I've said this before, but whatever. To me, this just screams Fourier transform. With the cosine, you have the real part of the inverse Fourier transform evaluated at t=1, with the original function being e^-2|t|, with its transform being 4/(w^2+4). By playing with the constants to make it for with the inverse Fourier transform, you get pi/2 time the original function at t=1, which is the result you got to via contouring.
A nice generalisation is: int_0^infty cos(1-1/x) \(x+1/x)^n dx becomes 2F0( {n/2,1-n/2},{}, -1/4) / (2^n*( n/2-1)!) for n>0 and an even integer. Also, int_0^infty sin(1-1/x) \(x+1/x)^n dx becomes 2F0( {n/2-1/2,3/2-n/2},{}, -1/4) / (2^n*( n/2-1/2)!) for n > 0 and an odd integer. (both equations multiplied by Pi/e^2)
Yes. The process is pretty much the same as in the problem given in the video. Completing the square in the denominator and applying the same u-sub will result in u^2+4.
Hello sir thank u so much for giving us such a quality content , could u post discussion regarding glasser master theorem i think it will be interesting .Thanks
You are free to choose the most convenient one. Usually you choose the contour so that it contains a single pole. And having convenient parametrization is also nice -- arcs are easily parametrized as z=rho e^(i phi) which simplifies estimation of the integral (you want any additional integrals equal zero). In fact, there is no simple answer to your question. You just try one contour and if it doesn't help you evaluate your integral you try another one. The more experience you have, the more often your guesses will be correct. Integrals are always just like this.
Because Contour usually for this case considers the geometry of the integral like in previous problem. So like in this form you see, you will get an answer that involve pi or e or something like that due to the radians and symmetry.
Gorgeous way to solve! Hail Kamal, thundermind of Lothlorien!!. Kamal, I think that I noticed in the first transformation from x to 1/x, you have the negative of the limit switch, another from the cos function symmetry, another from the differential element dx, is that correct?
Two substitutions u = 1/x and v = u-1/u gave me integral \int_{0}^{\infty}{\frac{cos(v)}{v^2+4}dx} which can be calculated fe with Laplace transform You did it for integral which looks almost the same ua-cam.com/video/bmZoPIfZLsw/v-deo.html
I=int[0,♾️](cos(x-1/x)/(x+1/x)^2)dx x-1/x=u x+1/x=sqrt(u^2+4) x=(u+sqrt(u^2+4))/2 du=(1+1/x^2)dx=(x+1/x)dx/x x->♾️, u->♾️ x->0, u->-♾️ I=1/2•int[-♾️,♾️]((u+sqrt(u^2+4))cos(u)/(u^2+4)^(3/2))du I1=1/2•int[-♾️,♾️](ucos(u)/(u^2+4)^(3/2))du I2=1/2•int[-♾️,♾️](cos(u)/(u^2+4))du I=I1+I2 U=cos(u) dV=(u/(u^2+4)^(3/2))du dU=-sin(u) V=-(u^2+4)^-1/2 I1=-1/2•int[-♾️,♾️](sin(u)/sqrt(u^2+4))du odd•even=odd, so I1=0 I=I2=1/2•int[-♾️,♾️](cos(u)/(u^2+4))du even•even=even, so I=int[0,♾️](cos(u)/(u^2+4))du b=u/2 db=du/2 I=1/2•int[0,♾️](cos(2b)/(b^2+1))db cos(x)=sum[n=0,♾️](x^(2n)/(2n)!) I=1/2•int[0,♾️](sum[n=0,♾️]((2b)^(2n)/((2n)!(b^2+1))))db I=sum[n=0,♾️](2^(2n-1)/(2n)!•int[0,♾️](b^(2n)/(b^2+1))db) O=b A=1 H=sqrt(b^2+1) tan(Ø)=b sec^2(Ø)=1/(b^2+1) sec^2(Ø)dØ=db I=sum[n=0,♾️](2^(2n-1)/(2n)!•int[0,pi/2](sin^2n(Ø)cos^-(2n+4)(Ø))dØ) int[0,pi/2](sin^(2u-1)(x)cos^(2v-1)(x))dx=ẞ(u,v) I=sum[n=0,♾️](2^(2n-1)ẞ(n+1/2,n+5/2)/(2n)!) ẞ(u,v)=gamma(u)gamma(v)/gamma(u+v) ẞ(n+1/2,n+5/2)=gamma(n+1/2)gamma(n+5/2)/gamma(2n+3) =(n+3/2)(n+1/2)gamma^2(n+1/2)/gamma(2n+3) I=pi/2•sum[n=1,♾️]((2n^2+n)(n-1)!!^2)/((2n)!^2)) I=pi/2•sum[n=0,♾️]((2n^2+5n+3)n!!^2/(2n+2)!^2) im not up to snuff on my double-factorial-squared sums, so I'll just watch the video now
One of the best integrals we'll watch YOU solve 😉
Hi,
"Terribly sorry about that" : 2:43 , 6:44 , 11:01 ,
"ok, cool" : 4:51 , 5:00 , 6:00 , 10:32 , 11:05 , 12:08 .
Guys i think he's terribly sorry about that
Hey, we did this in complex Analysis last year!!!! 😮
4:21 I feel like I've said this before, but whatever. To me, this just screams Fourier transform. With the cosine, you have the real part of the inverse Fourier transform evaluated at t=1, with the original function being e^-2|t|, with its transform being 4/(w^2+4). By playing with the constants to make it for with the inverse Fourier transform, you get pi/2 time the original function at t=1, which is the result you got to via contouring.
Like I said .....FOR MINAS TIRITH!!!!
@@maths_505 You get me there
@@maths_505 the greatest reason to solve that.
I see everyone proposed so many ways to reach the result but I think contour integration remains one of the best tools for solving hard integrals.
Hard integrals: (exist)
Contour integration: GO BACK TO THE SHADOWS!
FLAME OF UDÛN!!!
@@maths_505 it simply annihilate that seeming hardness as fire reduces wood to aches.
A nice generalisation is: int_0^infty cos(1-1/x) \(x+1/x)^n dx becomes 2F0( {n/2,1-n/2},{}, -1/4) / (2^n*( n/2-1)!) for n>0 and an even integer. Also, int_0^infty sin(1-1/x) \(x+1/x)^n dx becomes 2F0( {n/2-1/2,3/2-n/2},{}, -1/4) / (2^n*( n/2-1/2)!) for n > 0 and an odd integer. (both equations multiplied by Pi/e^2)
For the hw I find pi/(sqrt(2)*exp(sqrt(2))
Yes. The process is pretty much the same as in the problem given in the video. Completing the square in the denominator and applying the same u-sub will result in u^2+4.
After the cos(x)/x^2+4 part i stopped understanding what's happening but someday i do wanna learn about the integral for complex functions
could you do a video on how and why contour integration and the residue theorem works?
i would like to see such a video. I'm still unclear on what a laurent series is.
Answer 13:21 is pi/((sqrt(2)*e^(sqrt(2)))
Ingenious as always!
Answer for homework: pi/(2e^sqrt2)
Excellent
Hello sir thank u so much for giving us such a quality content , could u post discussion regarding glasser master theorem i think it will be interesting .Thanks
Thanks for smart tricks for solving such integrals. It should be possible to be solved in real domain rather than contour integration.
love me a good contour integral
How exactly does one choose the contour? Every time I see such integrals, I don’t understand why such a contour is chosen
You are free to choose the most convenient one. Usually you choose the contour so that it contains a single pole. And having convenient parametrization is also nice -- arcs are easily parametrized as z=rho e^(i phi) which simplifies estimation of the integral (you want any additional integrals equal zero). In fact, there is no simple answer to your question. You just try one contour and if it doesn't help you evaluate your integral you try another one. The more experience you have, the more often your guesses will be correct. Integrals are always just like this.
Because Contour usually for this case considers the geometry of the integral like in previous problem. So like in this form you see, you will get an answer that involve pi or e or something like that due to the radians and symmetry.
Gorgeous way to solve! Hail Kamal, thundermind of Lothlorien!!. Kamal, I think that I noticed in the first transformation from x to 1/x, you have the negative of the limit switch, another from the cos function symmetry, another from the differential element dx, is that correct?
The cos function is even so we don't get a negative from there my friend.
@@maths_505 thank you! I understood the opposite, my respect! Thundermind of Lothlorien!
What is the reason behind reversing the integral interval when changing the realm of x 1/x?
I have been trying to solve int^1_0 ln(1-x)/x dx without using the fact that ζ(2) = π^2/6. Is this possible?
Regarding 0:42, what exactly is the difference between a transformation as shown here and a regular substitution with u?
No difference whatsoever
Two substitutions
u = 1/x
and
v = u-1/u
gave me integral
\int_{0}^{\infty}{\frac{cos(v)}{v^2+4}dx}
which can be calculated fe with Laplace transform
You did it for integral which looks almost the same
ua-cam.com/video/bmZoPIfZLsw/v-deo.html
Wonderful professor, really wonderful. Please, professor, how can I reach this level of skill? Advise me what i should stady
Just keep doing math and you'll probably end up better than me
@@maths_505 Thank you sir and thank you for your humility Thank you, I hope that God will guide you to Islam
Why don't you do double and triple integrals? They are quite interesting!🎉
You must be new here
I've done plenty. One quite recently
@@maths_505 Do all of them
Bro may I know what books do you follow for complex analysis...like intregals including branch cuts and branch points ?
Complex analysis by gamelin is a classic
@@maths_505 thanks for replying
if im being completely honest i have no clue how to do complex analysis but I got the same result using laplace transform lol
The Laplace transform is a subject of complex analysis.
@@maths_505yea i meant the stuff you were doing with poles and contours, i learned laplace transform when studying differential equations
Fino al 5 Min è semplice...poi non conosco quelle soluzioni..il risultato arriva anche con feyman
I=int[0,♾️](cos(x-1/x)/(x+1/x)^2)dx
x-1/x=u
x+1/x=sqrt(u^2+4)
x=(u+sqrt(u^2+4))/2
du=(1+1/x^2)dx=(x+1/x)dx/x
x->♾️, u->♾️
x->0, u->-♾️
I=1/2•int[-♾️,♾️]((u+sqrt(u^2+4))cos(u)/(u^2+4)^(3/2))du
I1=1/2•int[-♾️,♾️](ucos(u)/(u^2+4)^(3/2))du
I2=1/2•int[-♾️,♾️](cos(u)/(u^2+4))du
I=I1+I2
U=cos(u)
dV=(u/(u^2+4)^(3/2))du
dU=-sin(u)
V=-(u^2+4)^-1/2
I1=-1/2•int[-♾️,♾️](sin(u)/sqrt(u^2+4))du
odd•even=odd, so I1=0
I=I2=1/2•int[-♾️,♾️](cos(u)/(u^2+4))du
even•even=even, so
I=int[0,♾️](cos(u)/(u^2+4))du
b=u/2
db=du/2
I=1/2•int[0,♾️](cos(2b)/(b^2+1))db
cos(x)=sum[n=0,♾️](x^(2n)/(2n)!)
I=1/2•int[0,♾️](sum[n=0,♾️]((2b)^(2n)/((2n)!(b^2+1))))db
I=sum[n=0,♾️](2^(2n-1)/(2n)!•int[0,♾️](b^(2n)/(b^2+1))db)
O=b
A=1
H=sqrt(b^2+1)
tan(Ø)=b
sec^2(Ø)=1/(b^2+1)
sec^2(Ø)dØ=db
I=sum[n=0,♾️](2^(2n-1)/(2n)!•int[0,pi/2](sin^2n(Ø)cos^-(2n+4)(Ø))dØ)
int[0,pi/2](sin^(2u-1)(x)cos^(2v-1)(x))dx=ẞ(u,v)
I=sum[n=0,♾️](2^(2n-1)ẞ(n+1/2,n+5/2)/(2n)!)
ẞ(u,v)=gamma(u)gamma(v)/gamma(u+v)
ẞ(n+1/2,n+5/2)=gamma(n+1/2)gamma(n+5/2)/gamma(2n+3)
=(n+3/2)(n+1/2)gamma^2(n+1/2)/gamma(2n+3)
I=pi/2•sum[n=1,♾️]((2n^2+n)(n-1)!!^2)/((2n)!^2))
I=pi/2•sum[n=0,♾️]((2n^2+5n+3)n!!^2/(2n+2)!^2)
im not up to snuff on my double-factorial-squared sums, so I'll just watch the video now
❤
29th
real okay cool moment
This video has 69 likes at the moment. I'm so conflicted...
Fool of you to think I'll solve this integrale