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While reading a textbook on Laplace transform, I found this sentence.

The inverse Laplace transform of $\mathcal{L}_X(s)=\left(\frac{\lambda_1}{s+\lambda_1}\right)\left(\frac{\lambda_2}{s+\lambda_2}\right)\cdots\left(\frac{\lambda_n}{s+\lambda_n}\right)$ is $$f_X(x) = \sum_{i=1}^n \alpha_i\lambda e^{-\lambda_i x}\quad\text{where}\quad \alpha_i = \prod_{\begin{smallmatrix}1\le j\le n\\ j\neq i\end{smallmatrix}}\frac{\lambda_j}{\lambda_j - \lambda_i}.$$

How can I prove it?

Note that the coefficients $\alpha_i$'s can be written as $\alpha_i = \ell_i(0)$ where each $\ell_i(x)$ is the Lagrange basis polynomial $$\ell_i(x) = \prod_{\begin{smallmatrix}1\le j\le n\\ j\neq i\end{smallmatrix}} \frac{x-x_j}{x_i-x_j} = \frac{(x-x_1)}{(x_i-x_1)} \cdots \frac{(x-x_{j-1})}{(x_i-x_{j-1})} \frac{(x-x_{j+1})}{(x_i-x_{j+1})} \cdots \frac{(x-x_n)}{(x_j-x_n)}$$ associated with points $\lambda_1,...,\lambda_n$.

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What happens if you take the Laplace transform of $f_X(x)$? –  J. M. Nov 27 '11 at 1:32

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We can write $$\left(\frac{\lambda_1}{s+\lambda_1}\right)\left(\frac{\lambda_2}{s+\lambda_2}\right)\cdots\left(\frac{\lambda_n}{s+\lambda_n}\right)\equiv\sum_{i=1}^n\frac{A_i}{s+\lambda_i}.$$ This implies that $$\lambda_1\cdots\lambda_n\equiv\sum_{i=1}^nA_i(s+\lambda_1)\cdots(s+\lambda_{i-1})(s+\lambda_{i+1})\cdots(s+\lambda_n).$$ Put $s=-\lambda_i$, we can find the coefficients $A_i$ in terms of $\lambda_j$ where $1\leq j\leq n$: $$A_i=\frac{\lambda_1\cdots\lambda_n}{(\lambda_1-\lambda_i)\cdots(\lambda_{i-1}-\lambda_i)(\lambda_{i+1}-\lambda_i)\cdots(\lambda_n-\lambda_i)}=\prod_{\begin{smallmatrix}1\le j\le n\\ j\neq i\end{smallmatrix}}\frac{\lambda_j}{\lambda_j - \lambda_i}:=\alpha_i\mbox{ for }1\leq i\leq n.$$ Hence, since the inverse Laplace transform is linear, we have $$L^{-1}\Big\{\left(\frac{\lambda_1}{s+\lambda_1}\right)\left(\frac{\lambda_2}{s+\lambda_2}\right)\cdots\left(\frac{\lambda_n}{s+\lambda_n}\right)\Big\}=\sum_{i=1}^n\alpha_iL^{-1}\Big\{\frac{1}{s+\lambda_i}\Big\}=\sum_{i=1}^n\alpha_ie^{-\lambda_ix},$$ as required.

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