= Solution
For Borel sets $A_1,\ldots,A_n$, multiply the finite-time <Radon-Nikodym derivative> into the product law of the increments:
$$
\begin{aligned}
\mathbb P_n^\lambda(X_1\in A_1,\ldots,X_n\in A_n)
&=\mathbb E\left[\prod_{j=1}^n\frac{e^{\lambda X_j}}{\phi(\lambda)}\mathbf1_{\{X_j\in A_j\}}\right]\\
&=\prod_{j=1}^n\int_{A_j}\frac{e^{\lambda x}}{\phi(\lambda)}\,\mu(dx).
\end{aligned}
$$
The factorization, extended from rectangles to the product <sigma-algebra>, proves <independence> and identical distributions. Their common law is the <exponential tilting>
$$
\boxed{\mu^\lambda(dx)=\frac{e^{\lambda x}}{\phi(\lambda)}\mu(dx).}
$$
For $\lambda>0$, $|x|e^{\lambda x}$ is bounded on $x\leq0$, while $xe^{\lambda x}\leq C_\varepsilon e^{(\lambda+\varepsilon)x}$ on $x\geq0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty)$ gives an integrable dominating function for the derivative. <Differentiation under the integral sign> then yields
$$
\phi'(\lambda)=\int xe^{\lambda x}\,\mu(dx),\qquad
\boxed{\mathbb E^\lambda X_1=\frac{\phi'(\lambda)}{\phi(\lambda)}\quad(\lambda>0).}
$$
The <mean under one-sided exponential tilting> needs an endpoint qualification. At $\lambda=0$, finiteness of positive <exponential moments> makes $X_1^+$ integrable, but gives no integrability of $X_1^-$. The formula holds with the right derivative and the extended mean $\phi'_+(0)=\mathbb EX_1\in[-\infty,\infty)$. To see this, apply <monotone convergence> to $(1-e^{-hX_1^-})/h$ and <dominated convergence> to $(e^{hX_1^+}-1)/h$ as $h\downarrow0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu(\{-j\})=6/(\pi^2j^2)$, $j\geq1$, has finite $\phi(\lambda)$ for every $\lambda\geq0$ but $\mathbb EX_1=-\infty$.
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