# Improper integral using comparison theorem

1. Oct 15, 2012

### markr2

State if the following integral converges or diverges, and justify your claim.

$\int_{-1}^{1} \frac{e^x}{x+1}\,dx$

I tried using the comparison theorem by comparing it to $\frac{1}{x+1}$. But for the interval (-1,0) the function is smaller for all x. So I could not conclude whether it diverged or converged.

If the comparison had concluded that the function was larger for all x I could have said it would diverge.

So my problem is finding the right equation to compare it to.

2. Oct 15, 2012

### tiny-tim

welcome to pf!

hi markr2! welcome to pf!
why 1 on top?

3. Oct 15, 2012

### markr2

Thanks for your reply, but I found the solution by comparing it to $\frac{e^{-1}}{x+1}$.

Should I delete this thread now?

4. Oct 15, 2012

### tiny-tim

no

5. Oct 15, 2012

### SammyS

Staff Emeritus

Besides your comparison is the wrong way around.

6. Oct 15, 2012

### Zondrina

Notice your integral has a discontinuity at -1.

$\int_{-1}^{1} \frac{e^x}{x+1} dx$

You either want something BIGGER that CONVERGES or you want something SMALLER that DIVERGES.

This integral looks like it's going to diverge to me, could you think of a smaller function which we can integrate that will DIVERGE here? That is :

$\int_{-1}^{1} \frac{e^x}{x+1}dx ≥ \int_{-1}^{1} something that diverges \space dx$

7. Oct 15, 2012

### SammyS

Staff Emeritus
Zondrina,

Use \text{} or \mbox{} to do text with spaces, etc. in Latex.

$\dots \int_{-1}^{1} \text{something that diverges} \space dx$