Laplace transform gives a wrong result when finding Renewal function In page 188 of the book Adventures in Stochastic Processes, it shows that if $F(dx) = xe^{-x}dx$, then the renewal function $U(x)$ will have the following expression $$U(x) = \frac{3}{4}+\frac{x}{2}+\frac{1}{4}e^{-2x}$$
Now, I want to try it with another approach. That is, I try to find the Laplace transform $\hat{F}$, then find the renewal function $U(x)$ by applying inverse Laplace transform on the renewal equation $\hat{U}(\lambda)=\frac{\hat{F^{0*}}(\lambda)}{1-\hat{F}(\lambda)}$.
My Approach
\begin{aligned} \hat{F} & =\int_0^\infty e^{-\lambda x} (xe^{-x})dx \\
&= \int_0^\infty xe^{-(\lambda+1)x}dx \\
&= \frac{1}{(\lambda+1)^2}\int_0^\infty ye^{-y}dy \ \ \ \ \text{(Let }y=(\lambda+1)x \text{)}\\
&=\frac{1}{(\lambda+1)^2}
\end{aligned}
Now, we also know that
\begin{aligned} 
\hat{U}(\lambda)&=\frac{1}{1-\hat{F}} \ \ \ \ \ \text{ (I remember that }\hat{F^{0*}}=1) \\
&=\frac{(\lambda+1)^2}{\lambda^2+2\lambda}\\
&=1+\frac{1}{\lambda^2+2\lambda}\\
&=1+\frac{1}{2\lambda}-\frac{1}{2(\lambda+2)}
\end{aligned}
Finally, applying inverse Laplace transform
\begin{aligned} 
U(x)&=\mathcal{L}^{-1}\{\hat{U}(\lambda)\}\\
&=\mathcal{L}^{-1}\{1+\frac{1}{2\lambda}-\frac{1}{2(\lambda+2)}\}\\
&=\delta(x)+\frac{1}{2}-\frac{1}{2}e^{-2x}
\end{aligned}
Which is different from the textbook
What's wrong here? Please correct me if I have any misconception
 A: Finally, I discovered that I've made a small mistake on the formula. It should be
$$\hat{U}=\frac{\hat{F^{0*}}}{\lambda(1-\hat{F})} \ \ \ \text{ (See Appendix B for the derivation)}$$
After having the correct formula, the steps are easy:
\begin{aligned} 
\hat{U}(\lambda)&=\frac{1}{\lambda(1-\hat{F})} \ \ \ \ \ (\hat{F^{0*}}=1 \text{ by definition}) \\
&=\frac{(\lambda+1)^2}{\lambda(\lambda^2+2\lambda)}\\
&=\frac{3}{4\lambda}+\frac{1}{2\lambda^2}+\frac{1}{4(\lambda+2)}
\end{aligned}
Now, by using the inverse of Laplace transform, the result will be
$$U(x) = \frac{3}{4}+\frac{x}{2}+\frac{1}{4}e^{-2x}$$
Appendix A - Method from the textbook
(I will just type out the exact wordings to prevent any mistakes)
We seek the density of of $\sum_{n=1}^\infty F^{n*}$. Observe that the Laplace transform of $\sum_{n=1}^\infty F^{n*}$ can be written as
$$
\begin{align}
\Bigg(\widehat{\sum_{n=1}^\infty F^{n*}}\Bigg)(\lambda) &=\sum_{n=1}^\infty \big(\widehat{F^{n*}}\big)(\lambda) = \sum_{n=1}^\infty\big(\hat{F}(\lambda) \big)^n \\
& = \sum_{n=1}^\infty ((1+\lambda)^{-2})^n \\
& = \frac{(1+\lambda)^{-2}}{1-(1+\lambda)^{-2}} = \frac{1}{1+2\lambda+\lambda^2-1} \\
& = \frac{1}{\lambda^2+2\lambda} = \frac{1}{\lambda(\lambda+2)}\\
\text{By a partial fraction expansion, this is}\\
&=\frac{1}{2\lambda}-\frac{1}{2(\lambda+2)} \\
&=\int_{x=0}^\infty e^{-\lambda x}\frac{1}{2}dx-\int_{x=0}^\infty e^{-\lambda x}\frac{1}{2}e^{-2x}dx\\
&=\int_{x=0}^\infty e^{-\lambda x}(\frac{1}{2}-\frac{1}{2}e^{-2x})dx \\
&=\int_{x=0}^\infty e^{-\lambda x}\Bigg(\sum_{n=1}^\infty F^{n*}\Bigg)dx
\end{align}
$$
Therefore, the required density is $\frac{1}{2}-\frac{1}{2}e^{-2x}$, since the transform uniquely determines the measure. Thus we have
$$
\begin{align}
U(x)&=\sum_{n=0}^{\infty}F^{n*}(x) \\
&= 1+\int_0^x\frac{1}{2}-\frac{1}{2}e^{-2s}ds\\
&= \frac{3}{4} + \frac{x}{2} +\frac{1}{4}e^{-2x}
\end{align}
$$
Appendix B - Proof for my formula
$$
\begin{align}
U(x)&=\sum_{n=0}^{\infty}F^{n*}(x) \\
&= F^{0*}+ \sum_{n=1}^{\infty}F^{n*}(x) \\
&= 1+ \sum_{n=1}^{\infty}F^{(n-1)*}*F(x) \\
&= 1+ \sum_{n=0}^{\infty}F^{n*}*F(x) \\
&= 1+ U*F(x)
\end{align}
$$
Now, applies Laplace transform on both side
$$
\begin{align}
\
\widehat{U}&=\frac{1}{\lambda}+\widehat{U}\widehat{F} \\
\widehat{U}(1-\widehat{F})&=\frac{1}{\lambda}\\
\widehat{U}&=\frac{1}{\lambda(1-\widehat{F})}
\end{align}
$$
