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Need help with simple improper integral

  • Thread starter Rib5
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  • #1
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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.

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

So I take the integral and get

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

gives me 0 - [tex]\frac{1}{-(b-s)}[/tex]

which is 1/(b-s)

The answer should be 1/(s-b). Can anyone help me figure out what I am messing up?
 

Answers and Replies

  • #2
Cyosis
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Where does the minus in your second step come from?
 
  • #3
gabbagabbahey
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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.

[tex]\int e^{t*(b-s)}[/tex] evaluated from 0 to [tex]\infty[/tex]
Click on the image below to see how to write this a little nicer with LaTeX:

[tex]\int_{0}^{\infty} e^{(b-s)t}dt[/tex]

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

So I take the integral and get

[tex]\frac{\int e^{(b-s)t}}{-(b-s)}[/tex] which evaluated from 0 to [tex]\infty[/tex]
Surely you mean

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

right?

Also, are you told that [itex](b-s)<0[/itex]? If not, you will need to examine two different cases.
 
  • #4
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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 [tex]e^{at}[/tex] is.

Also thanks for the tip about Latex
 

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