Integral function of the logarithm with exponential upper limit

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Let $F(x)=\int_1^{e^{x}}\ln t\,dt.$ (a) Verify, by integration by parts, that $G(t)=t\ln t-t$ is an antiderivative of $\ln t$, and compute $F(0)$ and $F(\ln 3)$. (b) Compute $F'(x)$ using the fundamental theorem of calculus. (c) Find the stationary points and, using $F''$, their nature.

Solution

(a) Antiderivative and values. Integrating by parts with $u=\ln t$, $dv=dt$: $\int\ln t\,dt=t\ln t-\int t\cdot\frac1t\,dt=t\ln t-t=G(t).$ Indeed $G'(t)=\ln t+t\cdot\frac1t-1=\ln t$. ✓

$F(0)=\int_1^{e^0}\ln t\,dt=\int_1^1\ln t\,dt=0.$ $F(\ln 3)=\int_1^{e^{\ln 3}}\ln t\,dt=\int_1^3\ln t\,dt=\big[t\ln t-t\big]_1^3=(3\ln 3-3)-(0-1)=3\ln 3-2\approx 1.2958.$

(b) Derivative. By the fundamental theorem with the chain rule, upper limit $e^x$: $F'(x)=\ln\big(e^{x}\big)\cdot e^{x}=x\,e^{x}.$

(c) Stationary points. Since $e^x>0$, $F'(x)=x\,e^{x}=0 \iff x=0.$ Differentiating: $F''(x)=e^x+x e^x=e^x(1+x)$, with $F''(0)=1>0$: so $x=0$ is a minimum (and $F(0)=0$).

$\boxed{F(0)=0,\quad F(\ln 3)=3\ln 3-2\approx 1.2958,\quad F'(x)=x e^{x},\quad x_{staz}=0\ (\text{minimum})}$