Question

Find all values for k for which the integral from 0 to 1 X^kLnX dx converges, and evalutae the integral for those values of k

Answer 1

\[\int\limits_{0}^{1} X^kLnX dx\]

Answer 2

integrated by parts and got to this point [lnx*x^(k+1)/(k+1)x^(k+1)/(k+1)^2] t, 1

Answer 3

[lnx*x^(k+1)/(k+1)-x^(k+1)/(k+1)^2] t, 1

Answer 4

I meant no k... nevermind that part

Answer 5

ok thanks

Answer 6

I think I have a partial solution
evaluate the answer I got above from 1 to 0 and turning the evaluation at zero into a right-hand limit you get\[\frac{-1}{(k+1)^2}-\lim_{n \rightarrow 0^+}\frac{1}{(k+1)^2}((k+1)n^{k+1}\ln n-n^{k+1})\]clearly n^(k+1) goes to zero as n does, and if \[\lim_{n \rightarrow 0^+}n^{k+1}\ln n=0\] as well this whole thing on the right is zero

I thinkthe limit can be done this way:\[\lim_{n \rightarrow 0}n^{k+1}\ln n=\lim_{t \rightarrow \infty}(\frac1t)^{k+1}\ln (\frac1t^{k+1})\]using the log property and remembering that k+1 is constant\[=(k+1)\lim_{t \rightarrow \infty}\frac{\ln\frac1t}{t^{k+1}}\]now we can use l'hospital's rule\[=\lim_{t \rightarrow \infty}-\frac1{t^{k+1}}=0\]that leaves us with\[\int_{0}^{1} x^k\ln xdx=\frac{-1}{(k+1)^2}\]so k cannot be -1
hm... there must be more, but this is a start
see you later, I'll keep thinking

Answer 7

thanks for your help.. imreally confuse about his problem

Answer 8

Having been thinking about it for a bit, I think the first answer above is above is very close to the actual solution.
I'll run through it again to try to make it more clear, then I will conclude with what I think is the full answer.
\[\int_{0}^{1}x^k\ln xdx=\frac{x^{k+1}}{(k+1)^2}[(k+1)\ln x-1]|_{0}^{1}\]because we have a singularity at x=0 we will have to convert this into a limit at that evaluation point, though evaluation at x=1 is no problem.
We get\[\int_{0}^{1}x^k\ln xdx=-\frac1{(k+1)^2}-\lim_{x \rightarrow 0 }\frac{x^{k+1}}{(k+1)^2}[(k+1)\ln x-1]\](actually it should be a right-hand limit, but I think I can show that the limit exists from both sides anyway so it's not necessary)

Just from looking at the solution we have so far we can conclude that\[k\neq-1\]Here is a part I missed last time: dealing with the limit we need to be a bit careful\[\lim_{x \rightarrow 0 }\frac{x^{k+1}}{(k+1)^2}[(k+1)\ln x-1]\]\[=\frac1{k+1}\lim_{x \rightarrow 0 }x^{k+1}\ln x-\frac1{(k+1)^2}\lim_{x \rightarrow 0 }x^{k+1}\]limit on the right is finite only if the exponent is non-negative\[k+1\ge0 \to k\ge-1\]and we already know that k can't be -1 because of the denominator, so we now only have to consider the limit when k>-1.
the second limit when k>-1 is\[\lim_{x \rightarrow 0 }x^{k+1}=0\]and for the first we can rearrange and use l'Hospital's rule\[\lim_{x \rightarrow 0 }x^{k+1}\ln x=\lim_{x \rightarrow 0 }\frac{\ln x}{\frac1{x^{k+1}}}=-\frac1{k+1}\lim_{x \rightarrow 0 }\frac{\frac1x}{\frac1{x^{k+2}}}\]\[=-\frac1{k+1}\lim_{x \rightarrow 0 }x^{k+1}=0\]putting it all together we find that\[\int_{0}^{1}x^k\ln xdx=-\frac1{(k+1)^2};k>-1\]and the integral diverges for all other values of k.

Answer 9

thanks turing you been a great help!