What is the Limit of the Natural Logarithm of Infinity Minus One?

[autobot.d]

\int^{\infty}_{1}\frac{1}{e^{t}-1}dt

= -ln(e - 1) + 1

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


Any help appreciated, thanks.

 

[CompuChip]

What anti-derivative did you find?
What is the limit as t \to \infty?

 

[DonAntonio]

\int^{\infty}_{1}\frac{1}{e^{t}-1}dt

= -ln(e - 1) + 1

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


Any help appreciated, thanks.

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

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

\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) , after evaluating the limit in infinity...

DonAntonio

 

[scurty]

\int^{\infty}_{1}\frac{1}{e^{t}-1}dt

= -ln(e - 1) + 1

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


Any help appreciated, thanks.

Eventually your answer will boil down to \lim\limits_{a \rightarrow \infty} (-ln|a| + ln|a-1| + ...) (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?