GATE CSE
First time here? Checkout the FAQ!
x
+1 vote
193 views
Evaluate the following definite integral :

$\int \limits_0^1 \log \left(\frac{1}{x} - 1 \right)$
asked in Calculus by Veteran (24.3k points)   | 193 views
Apply integration by parts,

it will be evaluated to 0.
Yes, answer is 0, but i didn't get it.

Let me try once more.
by parts we r getting $x log(1/x-1)+log(1-x) =0$

Your 2nd term is wrong.

it is -ln(1 - x)

why r u trying with ln, try with simple log
still no difference
yes, my mistake :(

2 Answers

+2 votes
Best answer
Applying the property of logarithms:

$log(\frac{1}{x}-1) = log(\frac{1-x}{x}) = log(1-x) - log(x)$

Integrating from 0 to 1 we can see that the same values are repeated in each term thus reducing the answer to zero.
answered by Junior (617 points)  
selected ago by
which property it is ?
Since log is a monotonically increasing function:

 $\int_{0}^{1} log(x) = \int_{0}^{1}log(1-x)$

and we know that log(a/b) = log(a) - log(b)
0 votes
ans is 0
still i want to cnfirm
xlg((1/x)-1)-log(1-x)
on putting x=0 /////     lg((1/x)-1) this will be undefined
answered by Junior (625 points)  
Top Users Feb 2017
  1. Arjun

    4898 Points

  2. Bikram

    4102 Points

  3. Habibkhan

    3748 Points

  4. Aboveallplayer

    2986 Points

  5. sriv_shubham

    2288 Points

  6. Smriti012

    2222 Points

  7. Arnabi

    1946 Points

  8. Debashish Deka

    1920 Points

  9. mcjoshi

    1614 Points

  10. sh!va

    1462 Points

Monthly Topper: Rs. 500 gift card

20,793 questions
25,951 answers
59,557 comments
21,976 users