How Do You Solve This Improper Integral with e^(t*(b-s))?

Rib5
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Hey guys, I was doing some homework problems and I ran into a problem regarding how to solve a certain improper integral.

\int e^{t*(b-s)} evaluated from 0 to \infty

So I take the integral and get

\frac{\int e^{t*(b-s)}}{-(b-s)} which evaluated from 0 to \infty

gives me 0 - \frac{1}{-(b-s)}

which is 1/(b-s)

The answer should be 1/(s-b). Can anyone help me figure out what I am messing up?
 
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Where does the minus in your second step come from?
 
Rib5 said:
Hey guys, I was doing some homework problems and I ran into a problem regarding how to solve a certain improper integral.

\int e^{t*(b-s)} evaluated from 0 to \infty

Click on the image below to see how to write this a little nicer with LaTeX:

\int_{0}^{\infty} e^{(b-s)t}dt

Is this what you meant? (you didn't actually specify which variable you are integrating over)

So I take the integral and get

\frac{\int e^{(b-s)t}}{-(b-s)} which evaluated from 0 to \infty

Surely you mean

\int_{0}^{\infty} e^{t*(b-s)}dt= \frac{e^{(b-s)t}}{(b-s)} {\left|}_{0}^{\infty}

right?

Also, are you told that (b-s)<0? If not, you will need to examine two different cases.
 
Thanks guys, I feel really stupid now. Earlier today I did a bunch of integrals where the sign on the power was negative and I think I ended up mixing up the what the integral of e^{at} is.

Also thanks for the tip about Latex
 
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