# Jump!

Problem Given $f(x) = e^x + \int\limits_{0}^{x} (t-x)f(t)\;dt\quad\quad\quad(\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

From $(\star)$, we see that $f(0) = 1;$ $f(x) = e^x + \int\limits_{0}^{x} t\cdot f(t) - x\cdot f(t) \;dt = e^x + \int\limits_{0}^{x} t\cdot f(t)\;dt-x\cdot \int\limits_{0}^{x}f(t)\;dt$.

And so, $\frac{df(x)}{dx}=\frac{de^x}{dx} + \frac{d}{dx}\int\limits_{0}^{x}tf(t)\;dt - \frac{d}{dx}\left(x\cdot \int\limits_{0}^{x}f(t)\;dt\right)$ $=e^x+\frac{d}{dx}\int\limits_{0}^{x}tf(t)\;dt-\left(\int\limits_{0}^{x}f(t)\;dt + x\frac{d}{dx}\int\limits_{0}^{x}f(t)\;dt\right)$ $\overset{\textbf{FTC}}{=}e^x + xf(x) -\int\limits_{0}^{x} f(t)\;dt - xf(x)$ $= e^x - \int\limits_{0}^{x}f(t)\;dt$

That is, $\frac{d}{dx}f(x)= e^x - \int\limits_{0}^{x}f(t)\;dt\implies f'(0) = 1.$

It follows that $\frac{d^2f(x)}{dx^2}=\frac{d}{dx}\left(e^x-\int\limits_{0}^{x}f(t)\;dt\right)=e^x-\frac{d}{dx}\left(\int\limits_{0}^{x}f(t)\;dt\right)\overset{\textbf{FTC}}{=}e^x-f(x).$

Solving $\begin{cases} f''(x)=e^x-f(x) \\f(0)=1\\f'(0)=1 \end{cases}\quad\quad\quad(\star\star)$

gives $f(x) = \frac{1}{2}(\sin(x)+\cos(x)+e^x).$

Fig. 1

Notice the derivation of $(\star\star)$ can be simplified if rgwe generalized Leibniz’s Rule (GLR, see “Deriving Generalized Leibniz’s Integral Rule”) is applied: $\frac{df(x)}{dx} = e^x + \underline{\frac{d}{dx}\int\limits_{0}^{x}(t-x)f(t)\;dt}$ $\overset{\textbf{GLR}}{=} e^x + \underline{(x-x)f(x)\cdot x'-(0-x)f(0)\cdot 0' + \int\limits_{0}^{x}\frac{\partial}{\partial x}(t-x)f(t)\;dt}$ $= e^x + \underline{\int\limits_{0}^{x}\frac{\partial}{\partial x}(t-x)f(t)\;dt}$ $=e^x+\int\limits_{0}^{x}-1\cdot f(t)\;dt$ $= e^x-\int\limits_{0}^{x}f(t)\;dt$ $\implies \frac{d^2f(x)}{dx}=e^x-\frac{d}{dx}\int\limits_{0}^{x}f(t)\;dt\overset{\textbf{FTC}}{=}e^x-f(x).$

Fig.2 shows that Omega CAS explorer‘s Maxima engine is both FTC and GLR aware:

Fig. 2

Exercise-1 Given: $f(x) = \int\limits_{0}^{x}t\cdot f(x-t)\;dt+\sin(x)$

where $f(x)$ is a continuous function, find $f(x).$

hint: Let $u=x-t, t = x-u; t=0\implies u=x; t=x\implies u=0; \frac{du}{dt}=-1$; $f(x) = \int\limits_{x}^{0}(x-u)\cdot f(u)\cdot (-1)\;du + \sin(x)=\int\limits_{0}^{x}(x-u)f(u)\;du+\sin(x).$