Integration question

• May 5th 2010, 09:59 AM
bidii
Integration question
I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks
• May 5th 2010, 11:01 AM
ebaines
Sorry, but as written the integral evaluates to infinity, not 1: the value of $\displaystyle e^{-a(t-2)}$ for $\displaystyle t \to -\infty$ is $\displaystyle \infty$.
• May 5th 2010, 11:19 AM
tonio
Quote:

Originally Posted by bidii
I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks

For $\displaystyle a\neq 0$ , $\displaystyle \int\limits_{-\infty}^\infty ae^{-a(t-2)}\,dt=-[e^{-a(t-2)}]^\infty_{-\infty}=-\lim_{b\to\infty}e^{-a(b-2)}+\lim_{h\to -\infty}e^{-a(h-2)}$ . Let's now check each limit separatedly:

$\displaystyle \lim_{b\to\infty}e^{-a(b-2)}=\lim_{b\to\infty}\frac{1}{e^{a(b-2)}}$ $\displaystyle =\left\{\begin{array}{lc}0&,\,if\,\,a>0\\\infty&,\ ,if\,\,a<0\end{array}\right.$ , and likewise $\displaystyle \lim_{h\to -\infty}e^{-a(h-2)}=\lim_{h\to -\infty}\frac{1}{e^{a(h-2)}}$ $\displaystyle =\left\{\begin{array}{lc}\infty&,\,if\,\,a>0\\0&,\ ,if\,\,a<0\end{array}\right.$

According to the above the value of the improper integral can't be 1 at all, since the integral diverges for any value of $\displaystyle a$.

I don't understand what you mean by "the expected value of t is 6"...

Tonio