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The function is like: $$F(x)=\int_{0}^{+\infty}(\ln t) (t^{x-1} )(e^{-t}) dt $$ Prove the integration's uniform convergence on $(0,+\infty)$

Well, you may notice that this is the derivative of $\Gamma$ function. The part I feel hard is how to process the $\ln t$ in it? Whether or not could you give the whole process, Thank you a lot!

Zhen Zhang
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    You know that it is the derivative of gamma function, then rewrite $\Gamma'(x)$ as $$ \Gamma'(x)=\Gamma(x)\cdot\frac{\Gamma'(x)}{\Gamma(x)}=\Gamma(x)\cdot\psi(x), $$ where $\psi(x)$ is a digamma function. – Tunk-Fey Jun 30 '14 at 11:50

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The integral $F(x)$ converges uniformly for $x \in [a,b]$ with $b> a >0$.

Consider, separately, the integrals over intervals $[1,\infty)$ and $[0,1]$.

Note that for $x \in [a,b]$ we have $|\ln(t)t^{x-1}e^{-t}| \leq \ln(t)t^{b-1}e^{-t}$ and

$$\lim_{t \rightarrow \infty}\frac{(\ln t)t^{b-1}e^{-t}}{t^{-2}}= 0,$$

Hence, there exists $c>0$ and $K>0$ such that $(\ln t)t^{b-1}e^{-t}< ct^{-2}$ for $x \in [a,b] , and $

$$\int_{K}^{\infty} (\ln t)t^{x-1}e^{-t}dt < \int_{K}^{\infty} t^{-2}dt = \frac1{K}.$$

It follows by the $M$-test that the integral over $[1,\infty)$ converges uniformly for $x \in [a,b]$.

For $t \in (0,1]$ and $x \geq a > 0$,

$$|(\ln t)t^{x-1}e^{-t}|< -(\ln t)t^{a-1}.$$

Hence,

$$\int_{0}^{1} |(\ln t)t^{x-1}e^{-t}|dt < -\int_{0}^{1} (\ln t)t^{a-1}dt = \frac1{a^2}$$

and the integral over $[0,1]$ converges uniformly for $x \in [a,\infty).$

However, the integral $F(x)$ does not converge uniformly on the 0pen interval $(0,\infty).$ We have for any $\delta$ with $0 < \delta < 1$,

$$\left|\int_{0}^{\delta} (\ln t)t^{x-1}e^{-t}dt\right| > e^{-1}\int_{0}^{\delta} (-\ln t)t^{t-1}dt = e^{-1}\frac{\delta^x (1-x\ln \delta)}{x^2}.$$

Since, for fixed $\delta$

$$\lim_{x \rightarrow 0}\frac{\delta^x (1-x\ln \delta)}{x^2} = \infty$$

we can find $x_\delta$ such that

$$\left|\int_{0}^{\delta} (\ln t)t^{x_\delta-1}e^{-t}dt\right| > 1.$$

RRL
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  • Thank you a lot for so much effort. But this problem is originally a part of "Prove that $\Gamma$ function has arbitrary order derivative on the interval $(0,+\infty)$". So I try to first prove that $\Gamma '(x)$ is uniformly convergent on this interval. – Zhen Zhang Jul 01 '14 at 01:17
  • Cant delete previous comment. You just need uniform convergence on $[a,\infty)$ for all $a>0$ -- which I proved -- to take the derivative under the gamma function integral. – RRL Jul 01 '14 at 01:30
  • You don't need as strong a condition as uniform convergence on $(0,\infty)$ - which does not hold. – RRL Jul 01 '14 at 01:32
  • $\Gamma'(x) = F(x)$ since we proved uniform convergence of $F$ on an interval that contains $x$. – RRL Jul 01 '14 at 01:47
  • Thank You. Do you mean that although a is defined on open interval, we can use the closed one $[a,+\infty]$ to prove the uniform convergence on open one $[0,+\infty]$ as long as it stands for all $a>0$? – Zhen Zhang Jul 01 '14 at 01:49
  • Yes -- you have all you need. For any $x > 0$ then uniform convergence of $F$ on a bracketing interval guarantees that $F(x)$ is the derivative. Everything diverges at $x=0$ so we avoid that. – RRL Jul 01 '14 at 01:54