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What are the detailed steps of the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculus Question: Improper integral $int_0^inftyfraccos(2x+1)sqrt[3]xdx$Finding the integral of $int_-infty^inftye^$.Computing a double integral $int_-infty^inftyint_-infty^inftyfracf(t)1+(x+g(t))^2dt dx$How do I solve this improper integral: $int_-infty^infty e^-x^2-xdx$?Evaluate the improper integral $int_-infty^0 e^xtextsin, xdx$Evaluate the $int_0^infty xe^-xdx$.Evaluate $int_-infty^inftycos(x)operatornamesech(x)dx$$int_-infty^infty e^-2ax^2dx=fracsqrtpisqrt2a$Improper integral $int_1^inftyfracdxxsqrtx^2 - 1$Improper integral with a parameter $int_0^infty e^-cx^2sin(tx)dx$










1












$begingroup$


What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.










share|cite|improve this question











$endgroup$











  • $begingroup$
    after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
    $endgroup$
    – Faraday Pathak
    Feb 22 at 6:31







  • 2




    $begingroup$
    The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
    $endgroup$
    – Andrei
    Feb 22 at 6:35















1












$begingroup$


What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.










share|cite|improve this question











$endgroup$











  • $begingroup$
    after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
    $endgroup$
    – Faraday Pathak
    Feb 22 at 6:31







  • 2




    $begingroup$
    The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
    $endgroup$
    – Andrei
    Feb 22 at 6:35













1












1








1


0



$begingroup$


What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.










share|cite|improve this question











$endgroup$




What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.







calculus integration definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 25 at 11:45









Harry Peter

5,53411439




5,53411439










asked Feb 22 at 5:53









user598604user598604

62




62











  • $begingroup$
    after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
    $endgroup$
    – Faraday Pathak
    Feb 22 at 6:31







  • 2




    $begingroup$
    The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
    $endgroup$
    – Andrei
    Feb 22 at 6:35
















  • $begingroup$
    after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
    $endgroup$
    – Faraday Pathak
    Feb 22 at 6:31







  • 2




    $begingroup$
    The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
    $endgroup$
    – Andrei
    Feb 22 at 6:35















$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31





$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31





2




2




$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35




$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35










1 Answer
1






active

oldest

votes


















0












$begingroup$

put $x-(a+bc^2)=t$



then,



$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$



we know,



$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$



for your case put $omega =0$ and $a=dfrac12c^2$



you'll get



$I=csqrt2pi$ notery along



note:
in general



$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$






share|cite|improve this answer











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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    put $x-(a+bc^2)=t$



    then,



    $I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$



    we know,



    $mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$



    for your case put $omega =0$ and $a=dfrac12c^2$



    you'll get



    $I=csqrt2pi$ notery along



    note:
    in general



    $displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$






    share|cite|improve this answer











    $endgroup$

















      0












      $begingroup$

      put $x-(a+bc^2)=t$



      then,



      $I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$



      we know,



      $mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$



      for your case put $omega =0$ and $a=dfrac12c^2$



      you'll get



      $I=csqrt2pi$ notery along



      note:
      in general



      $displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$






      share|cite|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        put $x-(a+bc^2)=t$



        then,



        $I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$



        we know,



        $mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$



        for your case put $omega =0$ and $a=dfrac12c^2$



        you'll get



        $I=csqrt2pi$ notery along



        note:
        in general



        $displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$






        share|cite|improve this answer











        $endgroup$



        put $x-(a+bc^2)=t$



        then,



        $I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$



        we know,



        $mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$



        for your case put $omega =0$ and $a=dfrac12c^2$



        you'll get



        $I=csqrt2pi$ notery along



        note:
        in general



        $displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 26 at 8:07

























        answered Feb 22 at 6:46









        Faraday PathakFaraday Pathak

        1,155316




        1,155316



























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