how to find the integral of a generalization of the Gaussian: exp(-abs(x)^q/2)












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I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.



The problem is to prove that:



$$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
is normalized. I know how to take care of the absolute value of course:



$$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$



but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...










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    0












    $begingroup$


    I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.



    The problem is to prove that:



    $$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
    is normalized. I know how to take care of the absolute value of course:



    $$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$



    but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.



      The problem is to prove that:



      $$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
      is normalized. I know how to take care of the absolute value of course:



      $$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$



      but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...










      share|cite|improve this question











      $endgroup$




      I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.



      The problem is to prove that:



      $$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
      is normalized. I know how to take care of the absolute value of course:



      $$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$



      but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...







      calculus integration statistics






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      edited Jan 18 at 22:22









      idriskameni

      749321




      749321










      asked Jan 18 at 21:49









      adventuringrawadventuringraw

      103




      103






















          1 Answer
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          $begingroup$

          Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
          $$
          frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
          $$

          To finish off, use the definition of the Gamma function:
          $$
          Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
          $$






          share|cite|improve this answer









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          • $begingroup$
            well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
            $endgroup$
            – adventuringraw
            Jan 18 at 23:30












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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

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          active

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          0












          $begingroup$

          Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
          $$
          frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
          $$

          To finish off, use the definition of the Gamma function:
          $$
          Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
            $endgroup$
            – adventuringraw
            Jan 18 at 23:30
















          0












          $begingroup$

          Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
          $$
          frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
          $$

          To finish off, use the definition of the Gamma function:
          $$
          Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
            $endgroup$
            – adventuringraw
            Jan 18 at 23:30














          0












          0








          0





          $begingroup$

          Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
          $$
          frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
          $$

          To finish off, use the definition of the Gamma function:
          $$
          Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
          $$






          share|cite|improve this answer









          $endgroup$



          Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
          $$
          frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
          $$

          To finish off, use the definition of the Gamma function:
          $$
          Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 22:28









          grand_chatgrand_chat

          20.5k11327




          20.5k11327












          • $begingroup$
            well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
            $endgroup$
            – adventuringraw
            Jan 18 at 23:30


















          • $begingroup$
            well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
            $endgroup$
            – adventuringraw
            Jan 18 at 23:30
















          $begingroup$
          well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
          $endgroup$
          – adventuringraw
          Jan 18 at 23:30




          $begingroup$
          well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
          $endgroup$
          – adventuringraw
          Jan 18 at 23:30


















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