- #1

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[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]

I'm out of ideas

- Thread starter neom
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- #1

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[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]

I'm out of ideas

- #2

lavinia

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When in doubt try integration by parts

[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]

I'm out of ideas

- #3

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Yes, of course. Thanks a lot!

- #4

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

- #5

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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.

- #6

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

- #7

disregardthat

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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.

- #8

lavinia

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Integrating by parts gives

[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]

I'm out of ideas

[tex]\frac{e^x}{x+1}|^{0}_{-1}+ \int_{-1}^0\frac{e^x}{(x+1)^2}dx[/tex]

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

Last edited:

- #9

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If one part diverges, then the whole thing diverges

- #10

lurflurf

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^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

[tex]\int_{-1}^0\frac{e^x+a}{x+1}dx[/tex]

can converge only when a =-e

and in particular not when a=0

- #11

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Because ln(0) is slightly undefined?

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.

- #12

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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.

- #13

lavinia

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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.

The second term diverges so first term diverges.

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