
#1
Apr1012, 09:10 AM

P: 69

[itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}1}dt[/itex]
[itex]= ln(e  1) + 1 [/itex] Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e. [itex] ln(e^{\infty} 1) = ??? [/itex] Any help appreciated, thanks. 



#2
Apr1012, 09:50 AM

Sci Advisor
HW Helper
P: 4,301

What antiderivative did you find?
What is the limit as [itex]t \to \infty[/itex]? 



#3
Apr1012, 09:53 AM

P: 606

I don't know how you solve this integral, but I must make a substitution: [itex]e^t=u\Longrightarrow t=\log u \Longrightarrow dt=\frac{du}{u}[/itex] , so the integral becomes: [itex]\int^{\infty}_e \frac{du}{u(u1)}=\int^\infty_e\frac{du}{u1}\int^\infty_e\frac{du}{u}= 1\log (e1)[/itex] , after evaluating the limit in infinity... DonAntonio 



#4
Apr1012, 08:49 PM

P: 381

Finite Integral, not sure whyYou need to take this limit, can you think of a way to combine the two natural logarithms in order to do this? 


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