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

• neom
In summary, the conversation is about finding a way to show that the integral from -1 to 0 of (e^x)/(x+1)dx diverges. Suggestions are made to use integration by parts and the comparison test. The conversation also mentions a warm-up question about why the integral of 1/x from 0 to 1 diverges. It is suggested to use a lower bound for the integrand to show that the integral diverges. The conversation also discusses the possibility of the integral converging if a= -e-1, but it is concluded that it cannot converge if a=0.

#### neom

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

neom said:
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?

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.

neom said:
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?

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If one part diverges, then the whole thing diverges

^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

homeomorphic said:
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.

vikiabst said:
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.

The argument i was making was not that. The first term, is easily seen to diverge since the numerator is finite and the denominator blows up at -1. It diverges to +infinity. If the second term is finite then the sum must diverge. But since the second term is positive, if it diverges, then it also diverges to plus infinity so the sum of the two must diverge to plus infinity.