Finding Value of $k$ for which $$\int^{1}_{0}\frac{\ln(x)}{x^k}$$ have finite value.
What I tried:
Put $\ln(x)=u$ and $x=e^u$ and $dx=e^udu$ and changing limits
$$I =\int^{0}_{-\infty}u\cdot e^{u(1-k)}du$$
$$I =\frac{1}{1-k}\biggl(u\cdot e^{u(1-k)}\biggr)\Biggm|^{0}_{-\infty}-\frac{1}{(1-k)^2}\biggl(e^{u(1-k)}\biggr)\Biggm |^{0}_{-\infty}$$
From above integral , it is sure that $k\neq1$
I have a problem in finding value at $-\infty$. Could some Help me to solve it. Thanks
Your work is correct. For the last step, consider limits. After some simplification you'll find that $$I=\underset{t\to\infty}{\text{lim}}\int_{-t}^0ue^{u (1-k)}\, du=\frac{-1+\underset{t\to\infty}{\text{lim}}e^{(k-1)t}(1-(k-1)t)}{(k-1)^2}$$ Substituting $n=k-1$ in the limit gives $$\underset{t\to \infty }{\text{lim}}e^{n t} (1-n t)$$ This limit converges to $0$ for all $\Re(n)<0$ and otherwise diverges. Therefore, the integral converges for $\Re(k-1)<0$ which is $\Re(k)<1$. The value of the integral is $$I=-\frac{1}{(k-1)^2}$$
To finish off the problem, note that $e^{ax}$ is valid for $x=-\infty$ only provided $a\ge 0.$ In your case we've already ruled out $a=0$ via other means, so that you must have $a>0.$ Of course, since $x/e^x$ vanishes for large $x,$ the exponential terms in your integral vanish, only provided $1-k>0,$ or in other words that $$k<1.$$
