Improper integral using comparison theorem

[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.

 

[tiny-tim]

welcome to pf!

hi markr2! welcome to pf!

I tried using the comparison theorem by comparing it to \frac{1}{x+1}.

why 1 on top?

 

[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?

 

[tiny-tim]

no

 

[SammyS]

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

Should I delete this thread now?

Never delete a thread.

Besides your comparison is the wrong way around.

 

[STEMucator]

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

 

[SammyS]

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} \text{something that diverges} \space dx

Zondrina,

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

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