Show that $int_0^pi /2fracdxsqrtcosx=int_0^pi /2fracdxsqrtsinx=fracGamma^2(frac14)2sqrt2 pi$ [duplicate]

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  • Compute the integral $int_0^fracpi2fracmathrm d thetasqrtsin theta$

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I am interested in showing that:
$$int_0^pi /2fracdxsqrtcosx=int_0^pi /2fracdxsqrtsinx=fracGamma^2(frac14)2sqrt2 pi$$
I've come across this equality without anything else, would you please help me what I should do?




definite-integrals closed-form




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marked as duplicate by Robert Z, Community♦ Jul 18 at 5:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Using Gamma function.....
    – Durgesh Tiwari
    Jul 18 at 5:08










  • Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
    – Michal Dvořák
    Jul 18 at 5:13














up vote
1
down vote

favorite
1













This question already has an answer here:



  • Compute the integral $int_0^fracpi2fracmathrm d thetasqrtsin theta$

    3 answers



I am interested in showing that:
$$int_0^pi /2fracdxsqrtcosx=int_0^pi /2fracdxsqrtsinx=fracGamma^2(frac14)2sqrt2 pi$$
I've come across this equality without anything else, would you please help me what I should do?




definite-integrals closed-form




share|cite|improve this question











marked as duplicate by Robert Z, Community♦ Jul 18 at 5:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Using Gamma function.....
    – Durgesh Tiwari
    Jul 18 at 5:08










  • Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
    – Michal Dvořák
    Jul 18 at 5:13












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






This question already has an answer here:



  • Compute the integral $int_0^fracpi2fracmathrm d thetasqrtsin theta$

    3 answers



I am interested in showing that:
$$int_0^pi /2fracdxsqrtcosx=int_0^pi /2fracdxsqrtsinx=fracGamma^2(frac14)2sqrt2 pi$$
I've come across this equality without anything else, would you please help me what I should do?




definite-integrals closed-form




share|cite|improve this question












This question already has an answer here:



  • Compute the integral $int_0^fracpi2fracmathrm d thetasqrtsin theta$

    3 answers



I am interested in showing that:
$$int_0^pi /2fracdxsqrtcosx=int_0^pi /2fracdxsqrtsinx=fracGamma^2(frac14)2sqrt2 pi$$
I've come across this equality without anything else, would you please help me what I should do?





This question already has an answer here:



  • Compute the integral $int_0^fracpi2fracmathrm d thetasqrtsin theta$

    3 answers





definite-integrals closed-form





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 18 at 5:05









Michal Dvořák

49112




49112




marked as duplicate by Robert Z, Community♦ Jul 18 at 5:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Robert Z, Community♦ Jul 18 at 5:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • Using Gamma function.....
    – Durgesh Tiwari
    Jul 18 at 5:08










  • Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
    – Michal Dvořák
    Jul 18 at 5:13
















  • Using Gamma function.....
    – Durgesh Tiwari
    Jul 18 at 5:08










  • Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
    – Michal Dvořák
    Jul 18 at 5:13















Using Gamma function.....
– Durgesh Tiwari
Jul 18 at 5:08




Using Gamma function.....
– Durgesh Tiwari
Jul 18 at 5:08












Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
– Michal Dvořák
Jul 18 at 5:13




Could you explain how? I'm quite unsure how to transform $int_0^infty t^x-1e^-tdt$ into $int_0^pi /2fracdxsqrtsinx$...
– Michal Dvořák
Jul 18 at 5:13










1 Answer
1






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Using the substitution $u=cos x$ gives
$$int_0^pi/2fracdxsqrtcos x
=int_0^1fracduu^1/2(1-u^2)^1/2.$$
Now let $t=u^2$ to get
$$int_0^1fracduu^1/2(1-u^2)^1/2
=frac12int_0^1fracdtt^3/4(1-t)^1/2=frac12B(1/4,1/2)$$
where $B(x,y)$ is the Beta function.
In terms of the Gamma function,
$$frac12B(1/4,1/2)=fracGamma(1/4)Gamma(1/2)2Gamma(3/4)
=fracsqrtpiGamma(1/4)2Gamma(3/4).$$
From the identity
$$Gamma(x)Gamma(1-x)=fracpisinpi x$$
we get
$$Gamma(1/4)Gamma(3/4)=pisqrt2$$
and then
$$fracsqrtpiGamma(1/4)2Gamma(3/4)
=fracGamma(1/4)^22sqrt2pi.$$






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Using the substitution $u=cos x$ gives
    $$int_0^pi/2fracdxsqrtcos x
    =int_0^1fracduu^1/2(1-u^2)^1/2.$$
    Now let $t=u^2$ to get
    $$int_0^1fracduu^1/2(1-u^2)^1/2
    =frac12int_0^1fracdtt^3/4(1-t)^1/2=frac12B(1/4,1/2)$$
    where $B(x,y)$ is the Beta function.
    In terms of the Gamma function,
    $$frac12B(1/4,1/2)=fracGamma(1/4)Gamma(1/2)2Gamma(3/4)
    =fracsqrtpiGamma(1/4)2Gamma(3/4).$$
    From the identity
    $$Gamma(x)Gamma(1-x)=fracpisinpi x$$
    we get
    $$Gamma(1/4)Gamma(3/4)=pisqrt2$$
    and then
    $$fracsqrtpiGamma(1/4)2Gamma(3/4)
    =fracGamma(1/4)^22sqrt2pi.$$






    share|cite|improve this answer

























      up vote
      3
      down vote



      accepted










      Using the substitution $u=cos x$ gives
      $$int_0^pi/2fracdxsqrtcos x
      =int_0^1fracduu^1/2(1-u^2)^1/2.$$
      Now let $t=u^2$ to get
      $$int_0^1fracduu^1/2(1-u^2)^1/2
      =frac12int_0^1fracdtt^3/4(1-t)^1/2=frac12B(1/4,1/2)$$
      where $B(x,y)$ is the Beta function.
      In terms of the Gamma function,
      $$frac12B(1/4,1/2)=fracGamma(1/4)Gamma(1/2)2Gamma(3/4)
      =fracsqrtpiGamma(1/4)2Gamma(3/4).$$
      From the identity
      $$Gamma(x)Gamma(1-x)=fracpisinpi x$$
      we get
      $$Gamma(1/4)Gamma(3/4)=pisqrt2$$
      and then
      $$fracsqrtpiGamma(1/4)2Gamma(3/4)
      =fracGamma(1/4)^22sqrt2pi.$$






      share|cite|improve this answer























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Using the substitution $u=cos x$ gives
        $$int_0^pi/2fracdxsqrtcos x
        =int_0^1fracduu^1/2(1-u^2)^1/2.$$
        Now let $t=u^2$ to get
        $$int_0^1fracduu^1/2(1-u^2)^1/2
        =frac12int_0^1fracdtt^3/4(1-t)^1/2=frac12B(1/4,1/2)$$
        where $B(x,y)$ is the Beta function.
        In terms of the Gamma function,
        $$frac12B(1/4,1/2)=fracGamma(1/4)Gamma(1/2)2Gamma(3/4)
        =fracsqrtpiGamma(1/4)2Gamma(3/4).$$
        From the identity
        $$Gamma(x)Gamma(1-x)=fracpisinpi x$$
        we get
        $$Gamma(1/4)Gamma(3/4)=pisqrt2$$
        and then
        $$fracsqrtpiGamma(1/4)2Gamma(3/4)
        =fracGamma(1/4)^22sqrt2pi.$$






        share|cite|improve this answer













        Using the substitution $u=cos x$ gives
        $$int_0^pi/2fracdxsqrtcos x
        =int_0^1fracduu^1/2(1-u^2)^1/2.$$
        Now let $t=u^2$ to get
        $$int_0^1fracduu^1/2(1-u^2)^1/2
        =frac12int_0^1fracdtt^3/4(1-t)^1/2=frac12B(1/4,1/2)$$
        where $B(x,y)$ is the Beta function.
        In terms of the Gamma function,
        $$frac12B(1/4,1/2)=fracGamma(1/4)Gamma(1/2)2Gamma(3/4)
        =fracsqrtpiGamma(1/4)2Gamma(3/4).$$
        From the identity
        $$Gamma(x)Gamma(1-x)=fracpisinpi x$$
        we get
        $$Gamma(1/4)Gamma(3/4)=pisqrt2$$
        and then
        $$fracsqrtpiGamma(1/4)2Gamma(3/4)
        =fracGamma(1/4)^22sqrt2pi.$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 18 at 5:36









        Lord Shark the Unknown

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