$F(x) =e^x + \int_0^1 (t e^x + F(t) e^{-x}) \text{d}t$

Question

$$F(x) =e^x + \int_0^1 (t e^x + F(t) e^{-x}) \text{d}t$$

What is $F(x)$ ?

Please help solve!

Answer 1

HINT:

If $F(x)$ satisfies the integral equation

$$F(x)=e^x+\int_0^1 (te^x+F(t)e^{-x})\,dt \tag 1$$

then $F(x)=\frac32 e^{x}+Ke^{-x}$. Now, substitute this form into $(1)$ and arrive at an equation that for $K$.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Note that we can write $(1)$ as $$F(x)=\frac32 e^x +\left(\int_0^1 F(t)\,dt\right)\,e^{-x}$$Therefore, we can write $F(x)=\frac32 e^x +Ke^{-x}$. Substituting this form into $(1)$ yields$$\begin{align}F(x)&=\frac32 e^x +Ke^{-x}\\\\&=\frac32 e^x +e^{-x}\int_0^1 \left(\frac32 e^{t}+Ke^{-t}\right)\,dt\\\\& =\frac32 e^x+\left(\frac32(e-1)-K(e^{-1}-1)\right)e^{-x}\tag 2\end{align}$$Solving $(2)$ for $K$ reveals $$K=\frac32 e(e-1)$$and hence we find that $$F(x)=\frac32 e^x+\frac32 e(e-1)e^{-x}$$.

Answer 2

First of all, split the integral in the first trivial part:

$$\int_0^1 te^x\ \text{d}t = \frac{1}{2}e^x$$

I know you can solve this by your own, it's really more than elementary.

The second part is

$$\int_0^1 F(t)e^{-x}\ \text{d}t$$

Again, the $e^x$ term can be pulled out because it doesn't depend upon $t$. The rest is given by the fundamental theorem of calculus, since your $F(t)$ is unknown. I'll call its primitive as $G(t)$.

$$e^{-x}\int_0^1 F(t)\ \text{d}t = e^{-x}\left(G(t=1) - G(t=0)\right)$$

Finally the result:

$$\frac{1}{2}e^x + e^{-x}\left(G(1) - G(0)\right)$$