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

[Rib5]

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?

 

[Cyosis]

Where does the minus in your second step come from?

 

[gabbagabbahey]

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.

 

[Rib5]

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