# Hello, can anyone tell me how I can show that this

Hello, can anyone tell me how I can show that this diverges

$$\int_{-1}^0\frac{e^x}{x+1}dx$$

I'm out of ideas

lavinia
Gold Member

Hello, can anyone tell me how I can show that this diverges

$$\int_{-1}^0\frac{e^x}{x+1}dx$$

I'm out of ideas

When in doubt try integration by parts

Yes, of course. Thanks a lot!

No, it's not working, I really can't evaluate this integral. Is there another way to see that it diverges?

As a warm-up, why does the integral of 1/x from 0 to 1 diverge?

It's not too different from that. Maybe use the comparison test, since you can bound e^x from below by some number greater than zero.

Because $$\lim_{x\to0}\frac1{x}$$ diverges? I'm not sure. What's the comparison test?

disregardthat

Find a lower bound for the integrand on [-1,0], that is, a function that is less than your integrand on [1,0]. (technically the absolute value of the integrand, but in this case it is positive)

Hint: what is the minimal value of e^x on [-1,0] ?

Then you can use the comparison test to see whether the integral of the lower bound diverges.

lavinia
Gold Member

Hello, can anyone tell me how I can show that this diverges

$$\int_{-1}^0\frac{e^x}{x+1}dx$$

I'm out of ideas

Integrating by parts gives

$$\frac{e^x}{x+1}|^{0}_{-1}+ \int_{-1}^0\frac{e^x}{(x+1)^2}dx$$

The first term diverges. If the second term is finite then the whole thing diverges. What if the second term diverges?

Last edited:

If one part diverges, then the whole thing diverges

lurflurf
Homework Helper

^No a sum of two divergent terms can converge such as (1/x-1/x) which converges even though each part diverges. For this problem show that
$$\int_{-1}^0\frac{e^x+a}{x+1}dx$$
can converge only when a =-e-1
and in particular not when a=0

As a warm-up, why does the integral of 1/x from 0 to 1 diverge?

It's not too different from that. Maybe use the comparison test, since you can bound e^x from below by some number greater than zero.

Because ln(0) is slightly undefined?

integral -1 to 0 (e^x)/1+x>e^-2[integral -1 to 0 1/(1+x)]

The second term diverges so first term diverges.

lavinia