Integration using the Feynman Trick
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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
add a comment |
up vote
2
down vote
favorite
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
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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 at 1:54
Q the Platypus
2,754933
2,754933
asked Nov 7 at 1:46
DavidG
1,042516
1,042516
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56
add a comment |
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56
@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56
add a comment |
5 Answers
5
active
oldest
votes
up vote
2
down vote
accepted
Here are some that I encountered ($I_1$ and $I_2$ are by far my favourites):
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
|
show 3 more comments
up vote
3
down vote
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 ;)
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
add a comment |
up vote
1
down vote
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
add a comment |
up vote
1
down vote
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}
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 at 15:28
add a comment |
up vote
1
down vote
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
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here are some that I encountered ($I_1$ and $I_2$ are by far my favourites):
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
|
show 3 more comments
up vote
2
down vote
accepted
Here are some that I encountered ($I_1$ and $I_2$ are by far my favourites):
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
|
show 3 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here are some that I encountered ($I_1$ and $I_2$ are by far my favourites):
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
Here are some that I encountered ($I_1$ and $I_2$ are by far my favourites):
$$I_1=int_0^frac{pi}{2} ln(sec^2x +tan^4x)dx$$
$$I_2=int_0^infty frac{lnleft({1+x+x^2}right)}{1+x^2}dx$$
$$I_3=int_0^frac{pi}{2}ln(2+tan^2x)dx$$
$$I_4=int_0^infty frac{x-sin x}{x^3(x^2+4)} dx$$
$$I_5=int_0^frac{pi}{2}arcsinleft(frac{sin x}{sqrt 2}right)dx$$
$$I_6=int_0^frac{pi}{2} lnleft(frac{2+sin x}{2-sin x}right)dx$$
$$I_7=int_0^frac{pi}{2} frac{arctan(sin x)}{sin x}dx $$
$$I_8=int_0^1 frac{ln(1+x^3)}{1+x^2}dx $$
$$I_9=int_0^{infty} frac{x^{4/5}-x^{2/3}}{ln(x)(1+x^2)}dx$$
$$I_{10}=int_0^1 frac{ln(1+x)}{x(1+x^2)}dx$$
In case you struggle where to put that parameter, feel free to ask.
edited Nov 20 at 23:37
answered Nov 15 at 21:02
Zacky
3,2181336
3,2181336
1
I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/…
– DavidG
Nov 20 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
|
show 3 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 at 4:16
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 at 0:40
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 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 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 at 4:16
1
1
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 at 9:59
That is nice to hear! I will try to add more when I have time.
– Zacky
Nov 20 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 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 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 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 at 0:40
1
1
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
I finally got $I_1$ out! math.stackexchange.com/questions/3026362/…
– DavidG
Dec 4 at 23:46
|
show 3 more comments
up vote
3
down vote
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 ;)
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
add a comment |
up vote
3
down vote
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 ;)
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
add a comment |
up vote
3
down vote
up vote
3
down vote
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 ;)
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 at 2:29
clathratus
2,305324
2,305324
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
add a comment |
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - Legend!! thanks very much!
– DavidG
Nov 7 at 3:30
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
@clatharus - I've solved two of these already :-) math.stackexchange.com/questions/2950772/… math.stackexchange.com/questions/2966938/…
– DavidG
Nov 7 at 3:32
add a comment |
up vote
1
down vote
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
add a comment |
up vote
1
down vote
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
add a comment |
up vote
1
down vote
up vote
1
down vote
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!
answered Nov 7 at 1:51
Henry Lee
1,684218
1,684218
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
add a comment |
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
Yes, have solved that one! Thanks though :-)
– DavidG
Nov 7 at 1:52
1
1
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
How about this: math.stackexchange.com/questions/2918366/…
– Henry Lee
Nov 7 at 1:53
add a comment |
up vote
1
down vote
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}
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 at 15:28
add a comment |
up vote
1
down vote
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}
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 at 15:28
add a comment |
up vote
1
down vote
up vote
1
down vote
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}
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 at 10:50
FDP
4,47411221
4,47411221
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 at 15:28
add a comment |
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 at 15:28
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 at 1:00
Is the upper limit meant to be $frac{pi}{2}$
– DavidG
Nov 16 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 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 at 15:28
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up vote
1
down vote
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
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
add a comment |
up vote
1
down vote
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
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
add a comment |
up vote
1
down vote
up vote
1
down vote
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
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$
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 at 11:02
Kevin
5,410822
5,410822
add a comment |
add a comment |
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@Q the Platypus - thanks for the edit.
– DavidG
Nov 7 at 1:56