How to prove the existence of $\displaystyle\int_0^1 t^x\ln(t)\ \mathrm{d}t$ when $x >-1$?

Question

How do I prove the existence of $\displaystyle\int_0^1 t^x\ln(t)\,\mathrm dt$ when $x >-1 $?

I know the integral under consideration is the Riemann integral but I don't see how to do this particular problem. Thanks for your help.

Answer 1

Let $t=e^u$ then: $$\int_{0}^{1} {t^xlntdt}=\int_{-\infty}^{0} {ue^{ux}e^udu}=\int_{-\infty}^{0}ue^{(x+1)u}du={{(x+1)u-1}\over{(x+1)^2}}e^{(x+1)u}|_{-\infty}^{0}=-{1\over {(x+1)^2}}$$

Answer 2

I would tackle by parts directly. If $u=\ln t$ and $\mathrm dv=t^x$ we have $$ \int_0^1 t^x\ln t\mathrm dt=-\lim_{t\to 0}\ln t\frac{t^{x+1}}{x+1}-\frac{1}{x+1}\int_0^1t^{x}\mathrm dt\\ =-\frac{1}{(x+1)^2} $$ by virtue of the fact that for any $\epsilon>0$ $$ \lim_{t\to 0}t^\epsilon\ln t=0 $$ and the condition that $x>-1$.

Note what happens when $x=-1$, where we have $$ 2\int_0^1\frac{\ln t}{t}\mathrm dt=-\lim_{t\to 0}(\ln t)^2 $$ which does not exist.

Answer 3

$$\begin{align}I&=\int_0^1 t^x\ln t\,\mathrm dt\\\text{Let }-u&=\ln t\implies t=e^{-u}\implies\mathrm dt=-e^{-u}\mathrm du\\I&=\int_\infty^0e^{-ux}(-u)(-e^{-u})\mathrm du\\&=-\int_0^\infty ue^{-(x+1)u}\,\mathrm du=-\int_0^\infty ue^{-su}\,\mathrm du\qquad[\text{Taking }s=(x+1)]\\&=-\mathcal{L}(u)\\&=-\dfrac{1!}{s^2}\qquad\qquad\qquad\left[\because\mathcal{L}(t^n)=\dfrac{n!}{s^{n+1}}\right]\\&=-\dfrac1{(x+1)^2}\end{align}$$

I have used Laplace Transform to compute the integral, it is clear that the integral converges only if $x\neq-1$. This is equivalent to saying that the integral exists for $x>-1$