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 integrable 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 regular domain in the $x\alpha$ -plane, whence one can change the order of integration 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 $$I \;=\; \int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\,dx \qquad (a > 0,\; b > 0),$$ 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: