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Hello, can anyone tell me how I can show that this diverges
[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]
I'm out of ideas
[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]
I'm out of ideas
When in doubt try integration by partsHello, can anyone tell me how I can show that this diverges
[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]
I'm out of ideas
Integrating by parts givesHello, can anyone tell me how I can show that this diverges
[tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]
I'm out of ideas
Because ln(0) is slightly undefined?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.
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.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.