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$$\int_0^1x^\alpha(1-x)^\beta \ln x dx$$

When $x\to 0$:

$(1-x)^\beta=1-\beta x+o(x)$

$$\int_0^1x^\alpha(1-\beta x)\ln xdx=\int_0^1x^\alpha\ln xdx-\int_0^1\beta x^{\alpha+1}\ln xdx$$

If we integrate by parts:

$$\int x^\alpha\ln xdx=\frac{x^{\alpha+1}}{\alpha+1}\ln x-\int\frac{x^\alpha}{\alpha+1}dx$$ $$\beta\int x^{\alpha+1}\ln xdx=\beta\frac{x^{\alpha+2}}{\alpha+2}\ln x-\beta\int\frac{x^{\alpha+1}}{\alpha+2}dx$$

So the integral converges when $\alpha>-1$, but in the answer it's written that $\beta>-2$. I don't see where they got that from. Are both conditions supposed to be met simultaneously or is one of them enough?

(By the way, I tried adding more members to the taylor expansion, but didn't really get to anything useful. Something tells me that that's where the answer lies but I can't how adding more members would affect the result.)

Belen
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3 Answers3

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Notice that when $\beta < 0$, the integral is also improper at the upper endpoint. It can be helpful to split the interval of integration at $1/2$. Then $\int_0^{1/2} \dots $ is analyzed as you have done. But $\int_{1/2}^1 \dots$ requires analysis for $\beta < 0$.

Eric Towers
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Remember that if $a>0$, then $x^a\log(x)\to0$ as $x\to0$ (you can check that easily). So if $a>0$ and $\beta\geq0$, the integral converges.

If $a\leq0$, then $\int_0^1x^a(1-x)^b\log(x)dx\geq\int_0^1(1-x)^b\log(x)dx\geq\int_0^{1/2}(1-x)^b\log(x)dx\geq c_b\int_0^{1/2}\log(x)dx=\infty$ for all $b\in\mathbb{R}$, where $c_b$ is a positive constant depending on $b$.

So what happens when $a>0$ and $b<0$? we have $$\int_0^1x^a(1-x)^b\log(x)dx=$$ $$=\int_0^{1/2}x^a(1-x)^b\log(x)dx+\int_{1/2}^1x^a(1-x)^bdx\sim \int_0^{1/2}x^a\log(x)dx+\int_{1/2}^1(1-x)^bdx$$ $\sim$ means that they behave in the same way with respect to convergence (I believe you can justify that yourself) So if $b<-1$, the integral converges, but if $b\in(-1,0)$, the integral diverges.

To sum up: if $a>0$ and $b\geq0$, we have convergence. If $a\leq0$, we have divergence (no matter how $b$ behaves). If $a>0$ and $b<-1$, we have convergence. If $a>0$ and $b\in(-1,0)$, we have divergence.

Just dropped in
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$$ \int_{0}^{1} x^{a}(1-x)^{b} \ln x \mathrm{~d} x=\int_{0}^{\frac{1}{2}} x^{a}(1-x)^{b} \ln x \mathrm{~d} x+\int_{\frac{1}{2}}^{1} x^{a}(1-x)^{b} \ln x \mathrm{~d} x :=I_{1}+I_{2} $$

For $I_1$, Consider $x\to 0^+$.Since $x$ is between $0$ and $\displaystyle\frac12$, $(1-x)^b$ does not have to participate in the discussion of convergence.

When $a>0$, we have $\displaystyle\lim _{x \rightarrow 0^{+}} x^{a}(1-x)^{b} \ln x=0$,so $I_1$ is convergent.

When $a=0$, easy to know $\displaystyle\int_{0}^{\frac{1}{2}}(1-x)^{b} \ln x \mathrm{~d} x$ and $$ \int_{0}^{\frac{1}{2}} \ln x \mathrm{~d} x=x \ln x\Bigg|_{0} ^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}} x \cdot \frac{1}{x} \mathrm{~d} x=\frac{1}{2} \ln \frac{1}{2}-\frac{1}{2} $$ have the same divergence.so $I_1$ is convergent.

When $-1<a<0$, take $\varepsilon$ satisfied $0<-a<\varepsilon<1$. At this time $\displaystyle\int_{0}^{\frac{1}{2}} \frac{1}{x^{\varepsilon}} \mathrm{d} x$ converges, and $$ \lim _{x \rightarrow 0^{+}} \frac{x^{a} \ln x}{\frac{1}{x^{\varepsilon}}} =\lim _{x \rightarrow 0^{+}} x^{a+\varepsilon} \ln x=0 $$ So $\displaystyle\int_{0}^{\frac{1}{2}} x^{a} \ln x \mathrm{~d} x$ converges, $I_1$ converges.

When $a \leqslant-1$, we have $$ \lim _{x \rightarrow 0^{+}} \frac{x^{a} \ln x}{\frac{1}{x}}=\lim _{x \rightarrow 0^{+}} x^{a+1} \ln x=\infty $$ Because $\displaystyle\int_{0}^{\frac{1}{2}} \frac{1}{x} \mathrm{~d} x$ diverges, so $\displaystyle\int_{0}^{\frac{1}{2}} x^{a} \ln x \mathrm{~d} x$ diverges, $I_1$ diverges.

For $I_2$, Consider $x\to 1^-$. $x^a$ does not have to participate in the discussion of convergence, and $$ \ln x=\ln (1+x-1) \sim x-1\qquad x \rightarrow 1^{-} $$ Therefore, it is equivalent to discussing the convergence of $$ -\int_{\frac{1}{2}}^{1}(1-x)^{b+1} \mathrm{~d} x=-\int_{\frac{1}{2}}^{1} \frac{1}{(1-x)^{-(b+1)}} \mathrm{d} x $$ When $-(b+1)<1$, that is $b>-2$, $I_2$ converges; when $b \leqslant -2$, we have $I_2$ diverges.

fairydeer
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