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integration under integral sign

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\newcommand{\sijoitus}[2]%
{\operatornamewithlimits{\Big/}_{\!\!\!#1}^{\,#2}}
\begin{document}
Let
$$I(\alpha) \;=\; \int_a^b\!f(x,\,\alpha)\,dx.$$
where \,$f(x,\,\alpha)$ is continuous in the rectangle
$$a \leqq x \leqq b,\, \quad \alpha_1 \leqq \alpha \leqq \alpha_2.$$
Then\, $\alpha \mapsto I(\alpha)$\, is continuous and hence \PMlinkname{integrable}{RiemannIntegrable} on the interval 
\,$\alpha_1 \leqq \alpha \leqq \alpha_2$;\, we have
$$\int_{\alpha_1}^{\alpha_2}I(\alpha)\,d\alpha \;=\; \int_{\alpha_1}^{\alpha_2}\left(\int_a^b\!f(x,\,\alpha)\,dx\right)d\alpha.$$
This is a double integral over a \PMlinkescapetext{regular domain} in the $x\alpha$-plane, whence one can change the \PMlinkname{order of integration}{FubinisTheorem} and accordingly write 
$$\int_{\alpha_1}^{\alpha_2}\left(\int_a^b\!f(x,\,\alpha)\,dx\right)d\alpha 
\;=\; \int_a^b\left(\int_{\alpha_1}^{\alpha_2}\!f(x,\,\alpha)\,d\alpha\right)dx.$$
Thus, a definite integral depending on a parametre may be integrated  with respect to this parametre by performing the integration under the integral sign.\\


\textbf{Example.}\, For being able to evaluate the improper integral
$$I \;=\; \int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\,dx \qquad (a > 0,\; b > 0),$$
we may interprete the integrand as a definite integral:
$$\frac{e^{-ax}-e^{-bx}}{x} \;=\; \sijoitus{\alpha=b}{\quad a}\!\frac{e^{-\alpha x}}{x} 
\;=\; \int_a^b\!e^{-\alpha x}\,d\alpha.$$
Accordingly, we can calculate as follows:
\begin{align*}
I & \;=\; \int_0^\infty\left(\int_a^b\!e^{-\alpha x}\,d\alpha\right)dx\\
& \;=\; \int_a^b\left(\int_0^\infty\!e^{-\alpha x}\,dx\right)d\alpha\\
& \;=\; \int_a^b\left(\sijoitus{x=0}{\quad\infty}\!-\frac{e^{-\alpha x}}{\alpha}\right)d\alpha\\
& \;=\; \int_a^b\!\frac{1}{\alpha}\,d\alpha \;=\; \sijoitus{a}{\quad b}\!\ln{\alpha}\\ 
& \;=\; \ln\frac{b}{a}
\end{align*}


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