Mixing
Integration using the Feynman Trick
I have a feeling this post won't met the community guidelines (will delete if so).
I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.
Does anyone know of any good ones to tackle?
integration
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
add a comment |
I have a feeling this post won't met the community guidelines (will delete if so).
I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.
Does anyone know of any good ones to tackle?
integration
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56
add a comment |
I have a feeling this post won't met the community guidelines (will delete if so).
I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.
Does anyone know of any good ones to tackle?
integration
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
I have a feeling this post won't met the community guidelines (will delete if so).
I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.
Does anyone know of any good ones to tackle?
integration
integration
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
edited Nov 7 '18 at 1:54
Q the Platypus
2,754933
2,754933
asked Nov 7 '18 at 1:46
DavidG
1,836619
asked Nov 7 '18 at 1:46
DavidG
1,836619
asked Nov 7 '18 at 1:46
DavidG
1,836619
1,836619
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56
add a comment |
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56
add a comment |
5 Answers
5
active
oldest
votes
Here are some that I have encountered:
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
answered Nov 15 '18 at 21:02
Zacky
5,0761752
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
|
show 5 more comments
A few good ones are:
$$int_0^infty e^{-frac{x^2}{y^2}-y^2}dx$$
$$int_0^infty frac{1-cos(xy)}xdx$$
$$int_0^infty frac{dx}{(x^2+p)^{n+1}}$$
$$int_{0}^{infty}e^{-x^2}dx$$
$$int_0^infty cos(x^2)dx$$
$$int_0^infty sin(x^2)dx$$
$$int_0^infty frac{sin^2x}{x^2(x^2+1)}dx$$
$$int_0^{pi/2} xcot x dx$$
That should keep you busy for a while ;)
answered Nov 7 '18 at 2:29
clathratus
3,243331
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
add a comment |
Maybe you can look at:
https://math.stackexchange.com/a/2989801/186817
Feynman's trick is used to compute:
begin{align}int_0^{frac{pi}{12}}ln(tan x),dxend{align}
answered Nov 8 '18 at 10:50
FDP
4,92911324
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
add a comment |
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
add a comment |
Another example is in evaluating
$$displaystyle int_0^infty dfrac{cos xdx}{1+x^2}$$
by first considering
$$Ileft(aright)=int_{0}^{infty}frac{sinleft(axright)}{xleft(1+x^{2}right)}dx,,a>0$$ we have $$I'left(aright)=int_{0}^{infty}frac{cosleft(axright)}{1+x^{2}}dx$$
From which it can be shown that
$$Ileft(aright)=frac{pi}{2}left(1-e^{-a}right)$$
hence
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
answered Nov 8 '18 at 11:02
Kevin
5,420822
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here are some that I have encountered:
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
answered Nov 15 '18 at 21:02
Zacky
5,0761752
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
|
show 5 more comments
Here are some that I have encountered:
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
answered Nov 15 '18 at 21:02
Zacky
5,0761752
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
|
show 5 more comments
Here are some that I have encountered:
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
answered Nov 15 '18 at 21:02
Zacky
5,0761752
Here are some that I have encountered:
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
answered Nov 15 '18 at 21:02
Zacky
5,0761752
edited Dec 23 '18 at 19:56
answered Nov 15 '18 at 21:02
Zacky
5,0761752
answered Nov 15 '18 at 21:02
Zacky
5,0761752
answered Nov 15 '18 at 21:02
Zacky
5,0761752
5,0761752
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
|
show 5 more comments
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
1
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 '18 at 4:16
1
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 '18 at 9:59
1
1
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/…
– DavidG
Dec 3 '18 at 23:27
1
1
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
That is nice to hear, as for $,I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx,$ I tried to keep it in the original way, but after the substitution $tan x =t$ might be clearer how to use Feynman's trick.
– Zacky
Dec 4 '18 at 0:40
1
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 '18 at 23:46
|
show 5 more comments
A few good ones are:
$$int_0^infty e^{-frac{x^2}{y^2}-y^2}dx$$
$$int_0^infty frac{1-cos(xy)}xdx$$
$$int_0^infty frac{dx}{(x^2+p)^{n+1}}$$
$$int_{0}^{infty}e^{-x^2}dx$$
$$int_0^infty cos(x^2)dx$$
$$int_0^infty sin(x^2)dx$$
$$int_0^infty frac{sin^2x}{x^2(x^2+1)}dx$$
$$int_0^{pi/2} xcot x dx$$
That should keep you busy for a while ;)
answered Nov 7 '18 at 2:29
clathratus
3,243331
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
add a comment |
A few good ones are:
$$int_0^infty e^{-frac{x^2}{y^2}-y^2}dx$$
$$int_0^infty frac{1-cos(xy)}xdx$$
$$int_0^infty frac{dx}{(x^2+p)^{n+1}}$$
$$int_{0}^{infty}e^{-x^2}dx$$
$$int_0^infty cos(x^2)dx$$
$$int_0^infty sin(x^2)dx$$
$$int_0^infty frac{sin^2x}{x^2(x^2+1)}dx$$
$$int_0^{pi/2} xcot x dx$$
That should keep you busy for a while ;)
answered Nov 7 '18 at 2:29
clathratus
3,243331
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
add a comment |
A few good ones are:
$$int_0^infty e^{-frac{x^2}{y^2}-y^2}dx$$
$$int_0^infty frac{1-cos(xy)}xdx$$
$$int_0^infty frac{dx}{(x^2+p)^{n+1}}$$
$$int_{0}^{infty}e^{-x^2}dx$$
$$int_0^infty cos(x^2)dx$$
$$int_0^infty sin(x^2)dx$$
$$int_0^infty frac{sin^2x}{x^2(x^2+1)}dx$$
$$int_0^{pi/2} xcot x dx$$
That should keep you busy for a while ;)
answered Nov 7 '18 at 2:29
clathratus
3,243331
A few good ones are:
$$int_0^infty e^{-frac{x^2}{y^2}-y^2}dx$$
$$int_0^infty frac{1-cos(xy)}xdx$$
$$int_0^infty frac{dx}{(x^2+p)^{n+1}}$$
$$int_{0}^{infty}e^{-x^2}dx$$
$$int_0^infty cos(x^2)dx$$
$$int_0^infty sin(x^2)dx$$
$$int_0^infty frac{sin^2x}{x^2(x^2+1)}dx$$
$$int_0^{pi/2} xcot x dx$$
That should keep you busy for a while ;)
answered Nov 7 '18 at 2:29
clathratus
3,243331
answered Nov 7 '18 at 2:29
clathratus
3,243331
answered Nov 7 '18 at 2:29
clathratus
3,243331
answered Nov 7 '18 at 2:29
clathratus
3,243331
3,243331
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
add a comment |
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 '18 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 '18 at 3:32
add a comment |
Maybe you can look at:
https://math.stackexchange.com/a/2989801/186817
Feynman's trick is used to compute:
begin{align}int_0^{frac{pi}{12}}ln(tan x),dxend{align}
answered Nov 8 '18 at 10:50
FDP
4,92911324
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
add a comment |
Maybe you can look at:
https://math.stackexchange.com/a/2989801/186817
Feynman's trick is used to compute:
begin{align}int_0^{frac{pi}{12}}ln(tan x),dxend{align}
answered Nov 8 '18 at 10:50
FDP
4,92911324
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
add a comment |
Maybe you can look at:
https://math.stackexchange.com/a/2989801/186817
Feynman's trick is used to compute:
begin{align}int_0^{frac{pi}{12}}ln(tan x),dxend{align}
answered Nov 8 '18 at 10:50
FDP
4,92911324
Maybe you can look at:
https://math.stackexchange.com/a/2989801/186817
Feynman's trick is used to compute:
begin{align}int_0^{frac{pi}{12}}ln(tan x),dxend{align}
answered Nov 8 '18 at 10:50
FDP
4,92911324
answered Nov 8 '18 at 10:50
FDP
4,92911324
answered Nov 8 '18 at 10:50
FDP
4,92911324
answered Nov 8 '18 at 10:50
FDP
4,92911324
4,92911324
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
add a comment |
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 '18 at 1:00
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
No, it's $dfrac{pi}{12}$. with upper bound to be $dfrac{pi}{2}$: begin{align}int_0^{frac{pi}{2}}ln(tan x),dx=0end{align} (perform the change of variable $y=dfrac{pi}{2}-x$ )
– FDP
Nov 16 '18 at 15:28
add a comment |
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
add a comment |
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
add a comment |
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
answered Nov 7 '18 at 1:51
Henry Lee
1,773218
1,773218
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
add a comment |
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 '18 at 1:52
1
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 '18 at 1:53
add a comment |
Another example is in evaluating
$$displaystyle int_0^infty dfrac{cos xdx}{1+x^2}$$
by first considering
$$Ileft(aright)=int_{0}^{infty}frac{sinleft(axright)}{xleft(1+x^{2}right)}dx,,a>0$$ we have $$I'left(aright)=int_{0}^{infty}frac{cosleft(axright)}{1+x^{2}}dx$$
From which it can be shown that
$$Ileft(aright)=frac{pi}{2}left(1-e^{-a}right)$$
hence
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
answered Nov 8 '18 at 11:02
Kevin
5,420822
add a comment |
Another example is in evaluating
$$displaystyle int_0^infty dfrac{cos xdx}{1+x^2}$$
by first considering
$$Ileft(aright)=int_{0}^{infty}frac{sinleft(axright)}{xleft(1+x^{2}right)}dx,,a>0$$ we have $$I'left(aright)=int_{0}^{infty}frac{cosleft(axright)}{1+x^{2}}dx$$
From which it can be shown that
$$Ileft(aright)=frac{pi}{2}left(1-e^{-a}right)$$
hence
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
answered Nov 8 '18 at 11:02
Kevin
5,420822
add a comment |
Another example is in evaluating
$$displaystyle int_0^infty dfrac{cos xdx}{1+x^2}$$
by first considering
$$Ileft(aright)=int_{0}^{infty}frac{sinleft(axright)}{xleft(1+x^{2}right)}dx,,a>0$$ we have $$I'left(aright)=int_{0}^{infty}frac{cosleft(axright)}{1+x^{2}}dx$$
From which it can be shown that
$$Ileft(aright)=frac{pi}{2}left(1-e^{-a}right)$$
hence
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
answered Nov 8 '18 at 11:02
Kevin
5,420822
Another example is in evaluating
$$displaystyle int_0^infty dfrac{cos xdx}{1+x^2}$$
by first considering
$$Ileft(aright)=int_{0}^{infty}frac{sinleft(axright)}{xleft(1+x^{2}right)}dx,,a>0$$ we have $$I'left(aright)=int_{0}^{infty}frac{cosleft(axright)}{1+x^{2}}dx$$
From which it can be shown that
$$Ileft(aright)=frac{pi}{2}left(1-e^{-a}right)$$
hence
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
answered Nov 8 '18 at 11:02
Kevin
5,420822
answered Nov 8 '18 at 11:02
Kevin
5,420822
answered Nov 8 '18 at 11:02
Kevin
5,420822
answered Nov 8 '18 at 11:02
Kevin
5,420822
5,420822
add a comment |
add a comment |
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@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 '18 at 1:56