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

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SUMMARY

The integral $$\int^1_0 \log(1-x) dx$$ can be evaluated using the limit approach and integration by parts. Despite the logarithm being undefined at 0, the limit $$\lim_{t \rightarrow 1} \int_{0}^{t} \ln(1-x) dx$$ allows for proper evaluation. This integral is classified as an improper integral, which requires careful handling to determine convergence or divergence.

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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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