Improper integral with logs

[dlthompson81]

Homework Statement

The problem is ∫e^x/(e^x)-1 to be evaluated from -1 to 1.

Homework Equations

The Attempt at a Solution

I got the integral as ln|(e^x)-1|

So, for the first part, evaluating from -1 to 0, with t being the limit at 0 I got this:

ln|(e^t)-1| - ln|(e^-1)-1|

which is supposed to evaluate to negative infinity, and I don't understand that

t is approaching 0, so e^t should be 1, 1-1 is 0, and ln0 doesn't exist and (e^-1)-1 is also negative which doesn't exist, so I'm a little confused as to how it works out to negative infinity.

Can anyone point me in the right direction? Thanks.

 

[Curious3141]

Homework Statement

The problem is ∫e^x/(e^x)-1 to be evaluated from -1 to 1.

Homework Equations

The Attempt at a Solution

I got the integral as ln|(e^x)-1|

So, for the first part, evaluating from -1 to 0, with t being the limit at 0 I got this:

ln|(e^t)-1| - ln|(e^-1)-1|

which is supposed to evaluate to negative infinity, and I don't understand that

t is approaching 0, so e^t should be 1, 1-1 is 0, and ln0 doesn't exist and (e^-1)-1 is also negative which doesn't exist, so I'm a little confused as to how it works out to negative infinity.

Can anyone point me in the right direction? Thanks.

What you want to do is to break up the integral into the sum of two parts like so:

lim (a → 0-) [ln (1 - e^a) - ln (1 - e^(-1))]

= lim (a → 0-) ln [(1 - e^a)/(1 - e^(-1))]

(the reason we swapped the terms around is because the modulus function applied on the argument for negative x necessitates a change of sign, otherwise the argument of the log function will be negative).

PLUS

lim (a → 0+) [ln (e^a - 1) - ln (e^1 - 1)]

= lim (a → 0+) ln [(e^a - 1)/(e^1 - 1)]

(here we don't need to swap the terms around because the argument is always positive).

Evaluate the two limits separately. Big hint: if you want to see why both ln(1 - e^a) and ln(e^a - 1) tend to negative infinity as a tends to zero, use the Taylor series for ln(z). which you can find on this Wiki page: http://en.wikipedia.org/wiki/Logarithm (search for "Taylor"). You'll find that at the limit, the series will become the negative harmonic series, which diverges to negative infinity. Since you're summing the two parts, the original improper integral also diverges to negative infinity.