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Suppose the improper integral $\displaystyle\int_{0}^{+\infty}xf(x)\,dx$ and $\displaystyle\int_{0}^{+\infty}\frac{f(x)}{x}\,dx$ are both convergent, prove that $$ I(t)=\int_{0}^{+\infty}x^tf(x)\,dx $$ is defined on $(-1,\,1)$ and has continuous derivative.

If $f(x)$ is nonnegative, then by comparison test, it's easy to prove it. But how to attack the general case?

Knt
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1 Answers1

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You may use Dirichlet's test in the following form: if $h(x)$ is such that $\left|\int_{0}^{t}h(x)\,dx\right| \leq C$ for any $t>0$ and $g(x)$ is such that $g(x)$ is decreasing to zero for any $x\geq x_0>0$, then $\int_{0}^{+\infty}h(x)g(x)\,dx$, as an improper Riemann integral, is convergent. Let us assume $t\in[0,1)$ and define $$ J(t) = \int_{0}^{+\infty} x^t \log(x) f(x)\,dx = \int_{0}^{+\infty}\underbrace{x f(x)}_{h(x)}\cdot\underbrace{\frac{\log x}{x^{1-t}}}_{g(x)}\,dx $$ and similarly $$ K(t) = \int_{0}^{+\infty} x^t \log^2(x) f(x)\,dx = \int_{0}^{+\infty}\underbrace{x f(x)}_{h(x)}\cdot\underbrace{\frac{\log^2 x}{x^{1-t}}}_{g(x)}\,dx. $$ By Dirichlet's test $K(t)$ is defined for any $t\in[0,1)$. By differentiation under the integral sign $K(t)=J'(t)$.
Similarly we get $J(t)=I'(t)$, so $I(t)$ is twice-differentiable (hence $C^1$) over $[0,1)$.
I will let you to deal with the case $t\in(-1,0)$, which is analogous.

Jack D'Aurizio
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  • I have a question. Like you said, $I(t)=\int_0^{\infty}xf(x)\frac{1}{x^{1-t}},dx$ given that $t\in[0,,1)$, then $\frac{1}{x^{1-t}}$ is undefined on $x=0$. I have tried Dirichlet's test, but it requires that $h(x),,g(x)$ are defined on $[0,,\infty)$. – Knt Aug 02 '19 at 15:00
  • The Riemann or Lebesgue integral $\int_{0}^{+\infty}f(x),dx$ are unchanged if $f(0)$ is re-defined, so it does not matter that the integrand function is not defined at the origin. – Jack D'Aurizio Aug 02 '19 at 15:24
  • I think you might not in the same mindset as me. I am still confused. To be clear, suppose $f(x)$ is continuous, now prove $I(t)$ is defined on $(-1,,1)$. For negative $t$, I mean that $\int_0^1 x^tf(x),dx$ may be an improper integral. I don't think it's just an affair of redefinition, for example, $\int_0^1\frac{1}{x},dx$ is divergent whatever value of $\frac{1}{x}$ you assume at point $0$. – Knt Aug 02 '19 at 17:26
  • @TaoX: in the same way, no matter that $\log(x)$ is undefined at the origin, $\int_{0}^{1}\log(x),dx = -1$. – Jack D'Aurizio Aug 02 '19 at 17:49