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?










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  • @Q the Platypus - thanks for the edit.
    – DavidG
    Nov 7 at 1:56















up vote
2
down vote

favorite
2












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?










share|cite|improve this question
























  • @Q the Platypus - thanks for the edit.
    – DavidG
    Nov 7 at 1:56













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





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?










share|cite|improve this question















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






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share|cite|improve this question













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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


















  • @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










5 Answers
5






active

oldest

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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.






share|cite|improve this answer



















  • 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


















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 ;)






share|cite|improve this answer





















  • @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


















up vote
1
down vote













you can try the most famous one which is:
$$int_0^inftyfrac{sin(x)}{x}dx$$
good luck!






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  • 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


















up vote
1
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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}






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  • 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


















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
$$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$






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    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.






    share|cite|improve this answer



















    • 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















    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.






    share|cite|improve this answer



















    • 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













    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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














    • 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










    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 ;)






    share|cite|improve this answer





















    • @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















    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 ;)






    share|cite|improve this answer





















    • @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













    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 ;)






    share|cite|improve this answer












    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 ;)







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • @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










    up vote
    1
    down vote













    you can try the most famous one which is:
    $$int_0^inftyfrac{sin(x)}{x}dx$$
    good luck!






    share|cite|improve this answer





















    • 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















    up vote
    1
    down vote













    you can try the most famous one which is:
    $$int_0^inftyfrac{sin(x)}{x}dx$$
    good luck!






    share|cite|improve this answer





















    • 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













    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!






    share|cite|improve this answer












    you can try the most famous one which is:
    $$int_0^inftyfrac{sin(x)}{x}dx$$
    good luck!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • 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










    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}






    share|cite|improve this answer





















    • 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















    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}






    share|cite|improve this answer





















    • 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













    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}






    share|cite|improve this answer












    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}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • 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










    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
    $$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$






    share|cite|improve this answer

























      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
      $$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$






      share|cite|improve this answer























        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
        $$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$






        share|cite|improve this answer












        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
        $$lim_{arightarrow1}I'left(aright)=frac{pi }{2e}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 8 at 11:02









        Kevin

        5,410822




        5,410822






























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