Solving a Problem from Reddit

This post used to be hosted on my GitHub, but I felt it would fit better here. I was intrigued by this problem because it tied into what I learned in Calculus BC. Only now do I know I didn't have the techniques I needed to tackle this problem. I can't quite put my finger on why, but my proof seems wrong to me. For instance, I prove an integral converges by bounding it between two values, which I can't immediately say is true. I'm not proud of this work, but I thought I'd put it here for completeness.

The Problem

Reddit user u/Poliorcetyks posed the following question on r/learnmath in his post titled Integrals with parameter, problem proving continuity. Given a function %% f(x) = \int_0^1 \frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt, %% find the domain of @@f@@ and show that it is continuous on its domain.

Domain

We begin by bounding @@f@@ between two other functions. Note that for all @@0 \leq t \leq 1@@ that %% \frac{1}{2} \leq \frac{1}{t+1} \leq 1 %% %% \frac{1}{2} \int_0^1 \ln\left(\frac{1}{t}\right)^x dt \leq f(x) \leq \int_0^1 \ln\left(\frac{1}{t}\right)^x dt, %% multiplying all sides of the inequality by the positive number @@\ln\left(1/t\right)^x@@ then integrating all sides from @@0@@ to @@1@@. Thus, we see that @@f@@ converges — and @@x@@ is in the domain of @@f@@ — if and only if the above integral also converges.

We can transform that integral by a @@u@@-substitution with \begin{align*} u &= \ln\left(\frac{1}{t}\right) = -\ln(t) \nl t &= e^{-u} \nl dt &= -e^{-u} du \end{align*} to get %% \int_0^1 \ln\left(\frac{1}{t}\right)^x dt = \int_0^\infty u^x e^{-u} du = \Pi(x). %% Thus, the above inequality simplifies to %% \frac{\Pi(x)}{2} \leq f(x) \leq \Pi(x) %% which is certainly less intimidating. All that remains is to determine the domain of @@\Pi@@, and we know that @@f@@ has the same domain.

It is clear that both @@0@@ and @@1@@ are in the domain of @@\Pi@@ as \begin{align*} \Pi(0) &= \int_0^\infty e^{-u} du = 1 \nl \Pi(1) &= \int_0^\infty ue^{-u} \nl &= \left[ -ue^{-u} + \int e^{-u} du \right]_0^\infty \nl &= 0 - 0 + 1 = 1. \end{align*} We will thus begin by showing that all @@0 < x < 1@@ are in the domain of @@\Pi@@. For those values of @@x@@, note that for @@0 \leq u \leq 1@@ that @@u^x \leq 1@@ and that for @@u > 1@@ that @@u^x < u@@. We can therefore, for @@0 < x < 1@@, split then bound @@\Pi@@ as \begin{align*} \Pi(x) &= \int_0^1 u^xe^{-u} du + \int_1^\infty u^xe^{-u} du \nl &\leq \int_0^1 e^{-u} du + \int_1^\infty ue^{-u} du \nl &\leq 1 + \frac{1}{e}. \end{align*} Since the integrand for @@\Pi@@ is always positive and since @@\Pi@@ is bounded for @@0 < x < 1@@, those values of @@x@@ are in the domain of @@\Pi@@ (and thus that of @@f@@).

We will now show by induction that all positive real numbers are in the domain of @@\Pi@@. We have already shown our base case that all positive real numbers less than @@1@@ are in the domain of @@\Pi@@. Now, assume that all positive real numbers less than @@n@@ are in the domain of @@\Pi@@. For any number @@n \leq x < n+1@@ we find that \begin{align*} \Pi(x) &= \left[ -u^xe^{-u} + \int xu^{x-1}e^{-u} \right]_0^\infty \nl &= 0 - 0 + x\,\Pi(x-1) = x\,\Pi(x-1) \end{align*} since @@x@@ is positive. Since @@x-1 < n@@ is in the domain of @@\Pi@@, all positive real numbers less than @@n+1@@ are in the domain of @@\Pi@@, proving by induction that all positive real numbers are in the domain of @@\Pi@@ and thus in the domain of @@f@@.

We will next show that all real numbers @@-1 < x < 0@@ are in the domain of @@\Pi@@. We perform an integration by parts on the definition of @@\Pi@@ as \begin{align*} \Pi(x) &= \left[ \frac{1}{x+1}\,u^{x+1}e^{-u} + \int \frac{1}{x+1}\,u^{x+1}e^{-u} \right]_0^\infty \nl &= 0 - 0 + \frac{\Pi(x+1)}{x+1} = \frac{\Pi(x+1)}{x+1} \end{align*} since @@x+1 > 0@@ and since @@x+1@@ is in the domain of @@\Pi@@ — and thus @@f@@ — as a result.

Note that @@x = -1@@ is not in the domain of @@\Pi@@. Attempting the integration by parts leads to \begin{align*} \Pi(-1) &= \left[ \ln(u)e^{-u} + \int \ln(u)e^{-u} du \right]_0^\infty \nl &= 0 + \infty + \int_0^\infty \ln(u)e^{-u} du \nl &= \infty \end{align*} since the integral in the last term is a convergent integral. There are similar results for all @@x < -1@@, as attempting the integration by parts gives %% \Pi(x) = \left[ \frac{1}{x+1}\,u^{x+1}e^{-u} + \int \frac{1}{x+1}\,u^{x+1}e^{-u} \right]_0^\infty %% which results in a division by zero since @@x + 1 < 0@@. (And of course, the integration by parts can continue upward until a convergent integral is reached. The result, however, is still the same.) Thus, no real numbers less than or equal to @@-1@@ are in the domain of @@\Pi@@, and are thus not in the domain of @@f@@ either.

In summary, we conclude that the domain of @@\Pi@@, and thus of @@f@@, is the set of all real numbers strictly greater than @@-1@@.

Continuity

To show that @@f@@ is continuous on its domain, we must show that @@\lim_{h \to 0} f(x+h) = f(x)@@. To show that, we first note two facts. First, that %% f(x) = \lim_{a \to 0^+} \int_a^{1-a} \frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt, %% since the integrand may go to infinity on either end — on the lower end if @@x < 0@@ and on the higher end if @@x > 0@@. And of course, it is the limit from the positive side since the integrand may not be defined outside the interval @@[0,1]@@. Second, that for all @@0 < t < 1@@ that %% \lim_{h \to 0} \ln\left(\frac{1}{t}\right)^h = 1. %%

With both of these facts in hand, the proof of continuity is remarkably simple. Note that %% \lim_{h \to 0} f(x+h) = \lim_{h \to 0} \int_0^1 \ln\left(\frac{1}{t}\right)^h \,\frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt. %% This limit exists if and only if for all @@\epsilon > 0@@ there exists a @@\delta > 0@@ such that when @@h@@ is within @@\delta@@ of @@0@@, @@f(x+h)@@ is within @@\epsilon@@ of @@f(x)@@. We use our first fact to see that there is an @@a > 0@@ such that %% \left| \int_a^{1-a} \frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt - f(x) \right| < \frac{\epsilon}{2}. %% We then use our second fact to show that for all @@h@@ within some distance @@\delta@@ of @@0@@, %% \left| \ln\left(\frac{1}{t}\right)^h - 1 \right| < \frac{\epsilon}{2 \left| \int_a^{1-a} \frac{\ln\left(\frac{1}{u}\right)^x}{u+1} \,du \right|} %% for all @@a < t < 1-a@@. Thus, we find that \begin{align*} \left| \int_a^{1-a} \ln\left(\frac{1}{t}\right)^h \frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt - \int_a^{1-a} \frac{\ln\left(\frac{1}{t}\right)^x}{t+1}\,dt \right| &= \left| \int_a^{1-a} \frac{\ln\left(\frac{1}{t}\right)^x}{t+1} \left( \ln\left(\frac{1}{t}\right)^h - 1 \right) \,dt \right| \nl &< \left|\frac{\epsilon}{2 \left| \int_a^{1-a} \frac{\ln\left(\frac{1}{u}\right)^x}{u+1} \,du \right|}\right| \left| \int_a^{1-a} \frac{\ln\left(\frac{1}{t}\right)^x}{t+1} \,dt \right| \nl &< \frac{\epsilon}{2}. \end{align*} Since it is possible to approximate @@f(x)@@, by choosing some @@a > 0@@ to cut our approximation off at, to within @@\epsilon/2@@, and since it is possible to choose an @@h@@ within @@\delta@@ of @@0@@ such that the first integral above is within @@\epsilon/2@@ of our approximation of @@f(x)@@, it follows that that integral is within @@\epsilon@@ of @@f(x)@@ as required. Thus, @@\lim_{h \to 0} f(x+h) = f(x)@@ and @@f@@ is continuous on its domain.