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Math 129-005H Homework 4 (improper integrals)

Kevin K. Lin

Due Wednesday 9/22

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1. Does

   $${\int }_0^{\mathrm{\infty }}{e}^{-t}𝑑t$$

   converge or diverge? If it converges, find its value.

   Solution:

   ${\displaystyle {\int }_0^b}{e}^{-t}𝑑t$ $=$ ${\left[-e^{-t}\right]|}_{t=0}^b$

   $=$ $-e^{-b}+e^0$

   $=$ $1-e^{-b}.$

   Since $e^{\mathrm{-}b}\mathrm{\to }\mathrm{0}$ as $b\mathrm{\to }\mathrm{\infty }$, the integral converges to $\mathrm{1}$.

2. Does

   $${\int }_0^{\mathrm{\infty }}t{e}^{-t}𝑑t$$

   converge or diverge? If it converges, find its value.

   Solution: integrate by parts with $u\mathrm{=}t$ and $v^{\mathrm{\prime }}\mathrm{=}{e}^{\mathrm{-}t}$:

   ${\displaystyle {\int }_0^b}t{e}^{-t}𝑑t$ $=$ ${\left[-t{e}^{-t}\right]|}_{t=0}^b+{\displaystyle {\int }_0^b}{e}^{-t}𝑑t$

   $=$ $-b{e}^{-b}+0+1-e^{-b}$

   Since $b\mathsf{}{e}^{\mathrm{-}b}\mathrm{\to }\mathrm{0}$ and $e^{\mathrm{-}b}\mathrm{\to }\mathrm{0}$ as $b\mathrm{\to }\mathrm{\infty }$, the integral converges to $\mathrm{1}$.

3. The gamma function is defined for all $x\ge 1$ by

   $$\mathrm{\Gamma }(x)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}𝑑t.$$

   It appears widely in many areas of mathematics and physics, for example in probability theory. (The function is actually defined for all $x>0$, but for it involves another kind of improper integral which we will discuss next week. )

   1. What are the values of $\mathrm{\Gamma }(1)$ and $\mathrm{\Gamma }(2)$? Hint: you’ve already done all the work above! No need to redo the work.

      Solution: $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{=}{\mathrm{\int }}_{\mathrm{0}}^{\mathrm{\infty }}{t}^{\mathrm{1}\mathrm{-}\mathrm{1}}\mathrm{\cdot }{e}^{\mathrm{-}t}\mathsf{}𝑑t\mathrm{=}\mathrm{1}$ and $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{2}\mathrm{)}\mathrm{=}{\mathrm{\int }}_{\mathrm{0}}^{\mathrm{\infty }}{t}^{\mathrm{2}\mathrm{-}\mathrm{1}}\mathrm{\cdot }{e}^{\mathrm{-}t}\mathsf{}𝑑t\mathrm{=}\mathrm{1}$

   2. Start with the definition of $\mathrm{\Gamma }(x+1)$ and integrate by parts once with respect to $t$ to find a simple relation between $\mathrm{\Gamma }(x+1)$ and $\mathrm{\Gamma }(x)$. This will lead to an expression for

      $$\frac{\mathrm{\Gamma }(x+1)}{\mathrm{\Gamma }(x)};$$

      what is it? (This problem may be a little confusing because there’s both $x$ and $t$. You should think of $x$ as a constant and treat $t$ as the variable.)

      Solution:

      Integrate by parts with $u\mathrm{=}{t}^x$, $v^{\mathrm{\prime }}\mathrm{=}{e}^{\mathrm{-}t}$:

      ${\displaystyle {\int }_0^b}{t}^x{e}^{-t}𝑑t$ $=$ $-{t^x{e}^{-t}|}_{t=0}^b+{\displaystyle {\int }_0^b}x{t}^{x-1}{e}^{-t}𝑑t$

      $=$ $-b^x{e}^{-b}+0+{\displaystyle {\int }_0^b}x{t}^{x-1}{e}^{-t}𝑑t$

      if $x\mathrm{>}\mathrm{0}$. As $b\mathrm{\to }\mathrm{\infty }$, $b^x\mathsf{}{e}^{\mathrm{-}b}\mathrm{\to }\mathrm{0}$, so

      $\mathrm{\Gamma }\left(x+1\right)$ $=$ ${\displaystyle {\int }_0^{\mathrm{\infty }}}x{t}^{x-1}{e}^{-t}𝑑t$

      $=$ $x{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{x-1}{e}^{-t}𝑑t\mathit{\hspace{1em}\hspace{1em}}\mathsf{\text{(because }}x\mathsf{\text{ is a “constant” here)}}$

      $=$ $x\mathrm{\Gamma }\left(x\right).$

      Thus $\mathrm{\Gamma }\mathsf{}\mathrm{(}x\mathrm{+}\mathrm{1}\mathrm{)}\mathrm{/}\mathrm{\Gamma }\mathsf{}\mathrm{(}x\mathrm{)}\mathrm{=}x$.

      (more space on next page)

   3. Using the above, find a simple expression for $\mathrm{\Gamma }\left(n\right)$. Hint: try calculating $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{3}\mathrm{)}$ in terms of $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{2}\mathrm{)}$, $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{4}\mathrm{)}$ in terms of $\mathrm{\Gamma }\mathsf{}\mathrm{(}\mathrm{3}\mathrm{)}$, etc. What pattern do you see?

      Solution: