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Finite Integral, not sure why

by autobot.d
Tags: finite, integral
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autobot.d
#1
Apr10-12, 09:10 AM
P: 68
[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1 [/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex] ln(e^{\infty} -1) = ??? [/itex]


Any help appreciated, thanks.
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CompuChip
#2
Apr10-12, 09:50 AM
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What anti-derivative did you find?
What is the limit as [itex]t \to \infty[/itex]?
DonAntonio
#3
Apr10-12, 09:53 AM
P: 606
Quote Quote by autobot.d View Post
[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1 [/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex] ln(e^{\infty} -1) = ??? [/itex]


Any help appreciated, thanks.


I don't know how you solve this integral, but I must make a substitution:

[itex]e^t=u\Longrightarrow t=\log u \Longrightarrow dt=\frac{du}{u}[/itex] , so the integral becomes:

[itex]\int^{\infty}_e \frac{du}{u(u-1)}=\int^\infty_e\frac{du}{u-1}-\int^\infty_e\frac{du}{u}= 1-\log (e-1)[/itex] , after evaluating the limit in infinity...

DonAntonio

scurty
#4
Apr10-12, 08:49 PM
P: 392
Finite Integral, not sure why

Quote Quote by autobot.d View Post
[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

[itex]= -ln(e - 1) + 1 [/itex]

Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
[itex] ln(e^{\infty} -1) = ??? [/itex]


Any help appreciated, thanks.
Eventually your answer will boil down to [itex]\lim\limits_{a \rightarrow \infty} (-ln|a| + ln|a-1| + ...)[/itex] (as you can see from DonAntonio's work above).

You need to take this limit, can you think of a way to combine the two natural logarithms in order to do this?


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