## Finite Integral, not sure why

$\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.
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 Quote by 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.

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

## Finite Integral, not sure why

 Quote by 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.
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?