How to show that for C , d &gt; 0, the integral ∫ 1 ∞ exp ⁡ ( − x d 2 C ) d x &lt; ∞

Question

How to show that for   C  ,  d  &amp;gt;  0, the integral      ∫    1          ∞        exp  ⁡  (  −            x      d              2      C        )  d  x  &amp;lt;  ∞

Blaine Foster · Accepted Answer

Note that by Taylor expansion of   exp function, for any   k  ≥  1 and   x  ≥  0 we have   exp  ⁡  (  x  )  ≥            x      k              k      !      . Hence for any   d  &amp;gt;  0,   C  &amp;gt;  0,   k  ≥  1 and   x  &amp;gt;  0 we get  exp  ⁡  (            x      d              2      C        )  ≥            (              x        d                    )        k                    (      2      C              )        k            k      !      Choose k such that   d  k  ≥  2. Let   M  =      1          (      2      C              )        k            k      !       (it&#039;s independent of x !). Then for   x  ≥  1 we get   exp  ⁡  (            x      d              2      C        )  ≥  M      x    2  . Hence      ∫    1          ∞        exp  ⁡  (  −            x      d              2      C        )  d  x  ≤      ∫          1        ∞        1          M              x        2              d  x  &amp;lt;  ∞

migongoniwt · Accepted Answer

Using the change of variable   u  =            x      d              2      C          ⟺    x  =            (      2      C      u      )                      1        d            , you have  I  =      ∫    1          ∞        exp  ⁡  (  −            x      d              2      C        )     d  x  =  A      ∫                  1                  2          C                            ∞            u                  1        d            −      1            e          −      u           d  xwhere A is a constant. And the integral of the RHS is convergent.

How to show that for C , d > 0 , the integral <msubsup> ∫

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