Question

Another fun integral!

Answer 1

sorry, keep doing what?

Answer 2

\[1=\int\limits_{0}^{\infty}te^{-tx}dx\] This should be true, check it yourself.


Then: \[t^{-1}=\int\limits_{0}^{\infty}te^{-tx}dx\] take the derivative of both sides with respect to t until you feel comfortable writing the formula for the n'th derivative of t.

Answer 3

Ahh that second equation should say:
\[t^{-1}=\int\limits_{0}^{\infty}e^{-tx}dx\]

Answer 4

right

Answer 5

So as an example, the first derivative of that eq wrt t is:
\[(-1)t^{-2}=\int\limits_{0}^{\infty}(-x)e^{-tx}dx\]