Antiderivative of x e − c x 2 I need to define c in ∫ 0 ∞ x e − c x 2 ,, so that it becomes a probability-mass function (so that it equals 1).Where do I even begin finding the antiderivative of this? I know the answer will be: e − c x 2 − 2 c .Trying to use partial integration: ∫ f ( x ) g ( x ) d x = f ( x ) G ( x ) − ∫ f ′ ( x ) G ( X )and picking x as my g(x), and e − c x 2 as my f(x) I end up with: f ′ ( x ) = − 2 c x f ( x ) G ( X ) = x 2 2 e − c x 2 x 2 2 − ∫ 0 ∞ − 2 c x f ( x ) x 2 2 d xWhich simplifies to: e − c x 2 x 2 2 − ∫ 0 ∞ − c x 4 e − c x 2 d x, which I find is just a mess.

Question

Antiderivative of   x      e          −      c              x        2            I need to define c in       ∫    0    ∞    x      e          −      c              x        2              ,, so that it becomes a probability-mass function (so that it equals 1).Where do I even begin finding the antiderivative of this? I know the answer will be:             e              −        c                  x          2                            −      2      c      .Trying to use partial integration:  ∫  f  (  x  )  g  (  x  )  d  x  =  f  (  x  )  G  (  x  )  −  ∫      f    ′    (  x  )  G  (  X  )and picking x as my g(x), and       e          −      c              x        2             as my f(x) I end up with:      f    ′    (  x  )  =  −  2  c  x  f  (  x  )  G  (  X  )  =            x      2        2        e          −      c              x        2                        x      2        2    −      ∫    0    ∞    −  2  c  x  f  (  x  )            x      2        2    d  xWhich simplifies to:       e          −      c              x        2                        x      2        2    −      ∫    0    ∞    −  c      x    4        e          −      c              x        2              d  x, which I find is just a mess.

Amir Beck · Accepted Answer

Step 1The antiderivative is easy. Note that ,since you want the integral to equal 1:                              ∫                      0                                ∞                          x                  e                      −            c                          x              2                                                        d                x        =        1                            ⇔                  1                      2            c                                    ∫                      0                                ∞                          2        c        x                  e                      −            c                          x              2                                                        d                x        =        1                                          ⇔                  1                      2            c                                                [            −                          e                              −                c                                  x                  2                                                      ]                    0                      ∞                          =        1                                          ⇔        2        c        =        1                                          ⇔        c        =        1                  /                2            Note also that c must be positive , otherwise the integral won&#039;t converge.Step 2Update: I fixed a major blunder. The quantity in brackets is evaluated as follows:      ∫          0              ∞        2  c  x      e          −      c              x        2                    d    x    =  −      lim          x      →      +      ∞            e          −      c              x        2              +      e    0    =  1(which is the definition of improper integral) and hence we go on to solve the equation.

glitinosim3 · Accepted Answer

Explanation:For   f  (  x  )  =      e          −      c              x        2             you have       f    ′    (  x  )  =  −  2  c  x  f  (  x  ), so integration by parts gives   ∫  x  f  (  x  )  d  x  =      1          −      2      c        ∫      f    ′    (  x  )  d  x  =      1          −      2      c        f  (  x  )  +  C.Now the definite integral.To have a probability-mass function, c has to be positive (otherwise the limit in infinity is intinite) and then   1  =  ∫  x  f  (  x  )  d  x  =      1          −      2      c        f  (  x  )            |        0          ∞        =      1          −      2      c        (  0  −  1  ), so   c  =      1    2   is the only solution.

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