integration under integral sign

Let

$$I(\alpha )={\int }_a^{b}f(x,\alpha )𝑑x.$$

where  $f(x,\alpha )$ is continuous in the rectangle

$$a\leqq x\leqq b,{\alpha }_1\leqq \alpha \leqq {\alpha }_2.$$

Then  $\alpha \mapsto I(\alpha )$  is continuous and hence integrable (http://planetmath.org/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 in the $x\alpha$-plane, whence one can change the order of integration (http://planetmath.org/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.

Example.  For being able to evaluate the improper integral

$$I={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}-e^{-bx}}{x}dx\mathit{\hspace{1em}\hspace{1em}}(a>0,b>0),$$

we may interprete the integrand as a definite integral:

$$\frac{e^{-ax}-e^{-bx}}{x}=\underset{\alpha =b}{\overset{a}{/}}\frac{e^{-\alpha x}}{x}={\int }_a^b{e}^{-\alpha x}𝑑\alpha .$$

Accordingly, we can calculate as follows:

$I$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle {\int }_a^b}{e}^{-\alpha x}𝑑\alpha \right)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_a^b}\left({\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\alpha x}𝑑x\right)𝑑\alpha$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_a^b}\left(\underset{x=0}{\overset{\mathrm{\infty }}{/}}-{\displaystyle \frac{e^{-\alpha x}}{\alpha }}\right)𝑑\alpha$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_a^b}{\displaystyle \frac{1}{\alpha }}𝑑\alpha =\underset{a}{\overset{b}{/}}\ln \alpha$

$\mathrm{\hspace{0.33em}}=\ln {\displaystyle \frac{b}{a}}$