MHB Integration of log(1-x) from 0 to 1

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To evaluate the integral of log(1-x) from 0 to 1, it's important to recognize that while log is undefined at 0, this does not prevent the evaluation of the integral. The integral is treated as an improper integral, requiring the limit approach as x approaches 1. Integration by parts can be employed to solve the integral effectively. The discussion emphasizes the need to compute the limit of the integral as the upper bound approaches 1. Understanding these concepts is crucial for correctly evaluating the integral.
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How to evaluate $$\int^1_0 log(1-x) dx $$

I am confused as log is not defined at 0. Please help
 
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suvadip said:
How to evaluate $$\int^1_0 log(1-x) dx $$

I am confused as log is not defined at 0. Please help

The fact that ln x is undefined for x=0 has no importance... what You have to do is computing... $\displaystyle \lim_{t \rightarrow 1} \int_{0}^{t} \ln (1-x)\ dx$

Kind regards

$\chi$ $\sigma$
 
suvadip said:
How to evaluate $$\int^1_0 log(1-x) dx $$

I am confused as log is not defined at 0. Please help

This is an improper integral which might converge or diverge. You can use integration by parts to solve it .
 
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