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Which of the following is true for $\int_{1}^{0} x\ln x\,\text dx$

  1. it is equal to $−1/4$
  2. it is divergent
  3. it is equal to an irrational number
  4. does not have a closed form
  5. it is impossible to evaluate this integral

According to my calculations it evaluates to $\frac{2x^{2}\ln x - x^{2}}{4}$
No when I put the values for limit 0, it then comes out to be $-\infty$ for $\log x$
Then which of the options are correct?

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    $\begingroup$ You should have $2\,x^2\,\ln x$ instead of $2\,\ln x$ $\endgroup$ Apr 14, 2015 at 21:53
  • $\begingroup$ After change of variable, it is $\int_0^{-\infty}xe^x\,dx$. $\endgroup$
    – Ziqian Xie
    Apr 14, 2015 at 22:00
  • $\begingroup$ @ZiqianXie: I believe it should be $\int_0^{-\infty}xe^{2x}\,\mathrm{d}x$ $\endgroup$
    – robjohn
    Apr 16, 2015 at 2:02
  • $\begingroup$ @robjohn yep, I forgot the $\frac{1}{4}$ factor. $\endgroup$
    – Ziqian Xie
    Apr 16, 2015 at 3:46
  • $\begingroup$ Which is correct from option 2, 3 and 4. $\endgroup$ Apr 17, 2015 at 2:36

1 Answer 1

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$\int_1^0 x\ ln(x)\ dx = \frac{x^2}{2}ln(x)|_1^0 - \int_1^0 \frac{x^2}{2}\frac{1}{x}dx= +\frac{1}{4}$

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  • $\begingroup$ which is correct from options 2 3 and 4 $\endgroup$ Apr 17, 2015 at 2:37
  • $\begingroup$ what would the correct option be if limits were from 0 to 1 $\endgroup$ Apr 17, 2015 at 2:42
  • $\begingroup$ That should be straightforward: $-\frac{1}{4}$ $\endgroup$ Apr 17, 2015 at 2:43

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