<s>[INST] Use Taylor Series Expansion in Calculating Integral I want to approximate this integral:
$$
I = \int_0^\infty  {{e^{ - bx}}\ln \left( {{a_1}{e^{ - {b_1}x}} + {a_2}{e^{ - {b_2}x}}} \right)dx} 
$$
where $b,{a_1},{a_2},{b_1},{b_2} > 0$ and $b_2>b_1$.
Here is my answer: 
We can observe that there exists a constant $c$ so that if $x>c$, then $\frac{{{a_2}}}{{{a_1}}}{e^{ - \left( {{b_2} - {b_1}} \right)x}} \in \left( {0,1} \right)$; and if $0 \le x \le c$, then $\frac{{{a_1}}}{{{a_2}}}{e^{ - \left( {{b_1} - {b_2}} \right)x}} \in \left( {0,1} \right]$. Actually the value of $c$ is: $$c = \frac{{\ln \left( {{a_1}/{a_2}} \right)}}{{{b_1} - {b_2}}}$$
Using the Taylor Series Expansion $\ln \left( {1 + y} \right) = \sum\limits_{n = 1}^\infty  {{{\left( { - 1} \right)}^{n - 1}}\frac{{{y^n}}}{n}}$ with $y \in \left( {0,1} \right)$, we obtain:
$$
\begin{array}{l}
I =  - \left( {\ln {a_1} - {b_1}} \right)\frac{{{e^{ - bc}}\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\int_c^\infty  {{e^{ - bx}}{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}{e^{ - n\left( {{b_2} - {b_1}} \right)x}}dx} } \\
 - \left( {\ln {a_2} - {b_2}} \right)\frac{{\left( {1 - {e^{ - bc}}} \right)\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\int_0^c {{e^{ - bx}}{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}{e^{ - n\left( {{b_1} - {b_2}} \right)x}}dx} } \\
 =  - \left( {\ln {a_1} - {b_1}} \right)\frac{{{e^{ - bc}}\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}\frac{{{e^{ - \left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}}} \\
 - \left( {\ln {a_2} - {b_2}} \right)\frac{{\left( {1 - {e^{ - bc}}} \right)\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}\frac{{1 - {e^{ - \left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \\
 =  - \frac{{\left( {b + 1} \right)}}{{{b^2}}}\left[ {\left( {\ln {a_1} - {b_1}} \right){e^{ - bc}} + \left( {\ln {a_2} - {b_2}} \right)\left( {1 - {e^{ - bc}}} \right)} \right]\\
 + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}\frac{{{e^{ - \left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}} + {{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}\frac{{1 - {e^{ - \left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \right)} 
\end{array}
$$
Since $\frac{{{a_2}}}{{{a_1}}}{e^{ - \left( {{b_2} - {b_1}} \right)c}} = \frac{{{a_1}}}{{{a_2}}}{e^{ - \left( {{b_1} - {b_2}} \right)c}} = 1$, we get
$$
\begin{array}{l}
I =  - \frac{{\left( {b + 1} \right)}}{{{b^2}}}\left[ {\left( {\ln {a_1} - {b_1}} \right){e^{ - bc}} + \left( {\ln {a_2} - {b_2}} \right)\left( {1 - {e^{ - bc}}} \right)} \right]\\
 + \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {\frac{{{e^{ - bc}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}} + \frac{{{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n} - {e^{ - bc}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \right)} 
\end{array}
$$
Is that result converged?
 [/INST] Substitution and repeated integration by parts yields
$$
\int_0^\infty x^ne^{-kx}\mathrm{d}x=\frac{n!}{k^{n+1}}
$$
Assume $b_1\le b_2$ and $|a_2|\le|a_1|$, but not both equal, then
$$
\begin{align}
&\int_0^\infty e^{-bx}\log\left(a_1e^{-b_1x}+a_2e^{-b_2x}\right)\mathrm{d}x\\
&=\int_0^\infty e^{-bx}\left(\log(a_1)-b_1x+\log\left(1+\frac{a_2}{a_1}e^{-(b_2-b_1)x}\right)\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\int_0^\infty e^{-bx}\log\left(1+\frac{a_2}{a_1}e^{-(b_2-b_1)x}\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\int_0^\infty\left(\color{#C00000}{\sum_{k=1}^\infty(-1)^{k-1}\frac1k\frac{a_2^k}{a_1^k}e^{-(b+k(b_2-b_1))x}}\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\sum_{k=1}^\infty(-1)^{k-1}\frac1k\frac{a_2^k}{a_1^k}\frac1{b+k(b_2-b_1)}\\
\end{align}
$$
where the series in red is derived from the Taylor series for $\log(1+x)$.
</s>