Differentiation Under the Integral Sign

Fix $x > 0$. On $t \in (0, 1]$ an antiderivative of $t^x$ is $\frac{t^{x+1}}{x+1}$, and $t^{x+1} \to 0$ as $t \to 0^+$ since $x + 1 > 0$.

Hence $F(x) = \int_0^1 t^x\,dt = \left[\frac{t^{x+1}}{x+1}\right]_0^1 = \frac{1}{x+1}$.

Differentiating in $x$ gives $F'(x) = -\frac{1}{(x+1)^2}$.

Now differentiate under the integral sign.

For $t \in (0, 1]$, $\frac{\partial}{\partial x} t^x = t^x \ln t$ (differentiate $t^x = e^{x \ln t}$).

At $t = 0$ define the integrand as $0$ — because $\lim_{t \to 0^+} t^x \ln t = 0$ for $x > 0$.

Compute $\int_0^1 t^x \ln t\,dt$ with $u = -(x+1)\ln t$. For $t \in (0, 1]$, $u \in [0, \infty)$, $t = e^{-u/(x+1)}$, $dt = -\frac{1}{x+1}e^{-u/(x+1)}\,du$.

Then $t^x \ln t\,dt = \frac{u e^{-u}}{(x+1)^2}\,du$, so $\int_0^1 t^x \ln t\,dt = -\frac{1}{(x+1)^2}\int_0^\infty u e^{-u}\,du$.

By parts, $\int_0^\infty u e^{-u}\,du = \left[-u e^{-u}\right]_0^\infty + \int_0^\infty e^{-u}\,du$.

The boundary term vanishes: it is $0$ at $u = 0$, and as $u \to \infty$, $e^u \ge \frac{u^2}{2}$ for $u \ge 0$ gives $0 \le u e^{-u} \le \frac{2}{u} \to 0$. Hence the integral is $0 + 1 = 1$; the under-integral derivative equals $-\frac{1}{(x+1)^2}$ — exactly the directly computed $F'(x)$.

Numerically: at $x = 1$ the analytic values are $F(1) = \frac12$, $F'(1) = -\frac14$. In the scene with $n = 200$ Simpson panels, $\int_0^1 t\,dt = 0.5$ exactly, and $\int_0^1 t\ln t\,dt$ lands within $2 \times 10^{-6}$ of $-\frac14$ — all three estimates converge to one point.

First, continuity. For $t > 0$, $t^x = e^{x\ln t}$ is continuous in $(t, x)$, so both $f$ and $f_x$ are continuous there.

At $t = 0$ we defined the value as $0$; for $x \ge \delta > 0$, $|t^x \ln t| = t^x |\ln t| \le t^{\delta}|\ln t| \to 0$ as $t \to 0^+$, uniformly in $x \in [\delta, A]$.

Hence $f$ and $f_x$ are continuous on the compact rectangle $[\delta, A] \times [0, 1]$.

Now prove the theorem: if $f$, $f_x$ are continuous on $[a, b] \times [c, d]$, then $F(x) = \int_c^d f(t, x)\,dt$ is differentiable with $F'(x_0) = \int_c^d f_x(t, x_0)\,dt$. The difference quotient is $\frac{F(x_0+h) - F(x_0)}{h} = \int_c^d \frac{f(t, x_0+h) - f(t, x_0)}{h}\,dt$.

Avoid selecting a mean-value point (its measurability in $t$ is not guaranteed) and use the integral form instead: since $f_x$ is continuous in $x$, $f(t, x_0+h) - f(t, x_0) = \int_{x_0}^{x_0+h} f_x(t, u)\,du = h \int_0^1 f_x(t, x_0 + \theta h)\,d\theta$, so $\frac{f(t, x_0+h) - f(t, x_0)}{h} - f_x(t, x_0) = \int_0^1 \left(f_x(t, x_0 + \theta h) - f_x(t, x_0)\right)\,d\theta$. Taking absolute values, $\left|\frac{f(t, x_0+h) - f(t, x_0)}{h} - f_x(t, x_0)\right| \le \sup_{|u - x_0| \le |h|} |f_x(t, u) - f_x(t, x_0)|$.

Since $f_x$ is uniformly continuous on the compact rectangle, this bound tends to $0$ uniformly in $t$ as $h \to 0$.

Hence $\left|\frac{F(x_0+h)-F(x_0)}{h} - \int_c^d f_x(t, x_0)\,dt\right| \le (d-c)\sup_t \sup_{|u - x_0| \le |h|} |f_x(t, u) - f_x(t, x_0)| \to 0$ — the limit-integral interchange is justified.

For any $x_0 > 0$, put $\delta = x_0/2$, $A = 2x_0$, so $x_0$ is interior to $[\delta, A]$.

Continuity from step 1 plus the theorem from step 2 give $F'(x_0) = \int_0^1 t^{x_0}\ln t\,dt$, which equals $-\frac{1}{(x_0+1)^2}$ by the direct computation of problem 1.

So the three estimates meeting in the scene is no accident — it is forced by uniform continuity on the compact rectangle.

Numerically: at $x = 2$ the analytic values are $F(2) = \frac13$, $F'(2) = -\frac19$.

With $n = 200$ the Simpson approximation of $\int_0^1 t^2 \ln t\,dt$ is within $10^{-8}$.

Even though $\ln t$ blows up near $t = 0$, the factor $t^2$ tames it ($t^2|\ln t| \to 0$), and the numerics show the integral converging cleanly.

First check $\int_0^1 \varphi = 1$: $\int_0^1 30u^2(1-u)^2\,du = 30\int_0^1 (u^2 - 2u^3 + u^4)\,du = 30\left(\frac13 - \frac12 + \frac15\right) = 30 \cdot \frac{1}{30} = 1$.

For $0 < x < 1$, $F(x) = \int_0^1 f(x, t)\,dt = \int_0^x 30(t/x)^2(1-t/x)^2\,dt$ (nonzero only for $0 < t < x$). With $u = t/x$, $F(x) = x\int_0^1 30u^2(1-u)^2\,du = x \cdot 1 = x$.

Since $F(0) = 0$, the right derivative at $0$ is $\lim_{h \to 0^+} \frac{F(h) - F(0)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.

Next, $\frac{\partial f}{\partial x}(0, t) = 0$. Fix $t > 0$: for $0 < h < t$ we have $t < h$ false, so $f(h, t) = 0$ ($f$ is nonzero only when $0 < t < h$).

Hence the difference quotient $\frac{f(h, t) - f(0, t)}{h} = 0$ for every $h < t$, and the limit is $0$. At $t = 0$, $f(h, 0) = 0$ too.

So $\frac{\partial f}{\partial x}(0, t) = 0$ for every $t \in [0, 1]$, and $\int_0^1 \frac{\partial f}{\partial x}(0, t)\,dt = 0 \ne 1$ — different from the right derivative. The under-integral formula breaks at $x = 0$.

Now locate the broken hypothesis. On the line $t = x/2$, $f(x, x/2) = 30\left(\frac12\right)^2\left(\frac12\right)^2 = \frac{15}{8} = 1.875$, independent of $x$; so $f \to 1.875 \ne 0 = f(0, 0)$ along $(x, x/2) \to (0, 0)$ — $f$ itself is discontinuous at $(0, 0)$.

The partial derivative is worse: with $u = t/x$, $\frac{\partial f}{\partial x} = -\frac{60u^2(1-u)(1-2u)}{x}$ ($0 < t < x$), and on the line $t = x/4$ ($u = 1/4$) this is $-\frac{45}{32x} \to -\infty$ as $x \to 0^+$.

So $f_x$ is unbounded in every neighborhood of $(0, 0)$, cannot be uniformly continuous, and the key step of problem 2's proof — uniform convergence of the difference quotients — collapses. Without the continuity hypotheses, the limit-integral interchange is not guaranteed — that is the moral of the counterexample.