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

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?

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

$\int_1^0 x\ ln(x)\ dx = \frac{x^2}{2}ln(x)|_1^0 - \int_1^0 \frac{x^2}{2}\frac{1}{x}dx= +\frac{1}{4}$
• That should be straightforward: $-\frac{1}{4}$ Apr 17, 2015 at 2:43