Improper integrals and convergence

Material by the IntegralPaso project. Conditions and limits are stated in each guide; external teaching review is pending.

An infinite endpoint or a singularity requires limits. For an interior singularity, split the integral there and require both sides to converge separately. A Cauchy principal value is not the same as a convergent improper integral. Finding a primitive does not establish convergence.

Worked example

1. Identify the integrand, variables and order. For multiple integrals, start with the innermost integral.

   $$\int\limits_{0}^{\infty} e^{- x}\, dx$$

2. Integrate with respect to x. Treat the other variables as constants at this stage.

   $$\int e^{- x}\, dx$$

3. Make a substitution and replace the differential too; both must change together.

   $$u = - x,\quad du = -1\,dx$$

4. Move the constant factor outside: it does not depend on the integration variable.

   $$\int \left(- e^{u}\right)\, du = - e^{u}$$

5. The exponential is its own derivative. Compensate any constant factor in the exponent.

   $$\int e^{u}\, du = e^{u}$$

6. Combine the results of the substeps and simplify.

   $$\int \left(- e^{u}\right)\, du = - e^{u}$$

7. After integrating the expression in u, return to the original variable.

   $$\int e^{- x}\, dx = - e^{- x}$$

8. Check this primitive by differentiating it: recover exactly the integrand of this stage.

   $$e^{- x} = e^{- x}$$

9. Replace the infinite endpoint with b and take the limit as b tends to infinity.

   $$\lim_{b \to \infty}\left(1 - e^{- b}\right) = 1$$

10. Apply the bounds: upper value minus lower value. For singularities or infinite bounds, the engine evaluates the integral as improper.

    $$\left[- e^{- x}\right]_{0}^{\infty} = 1$$

11. The result corresponds to the specified bounds and integration order.

    $$1$$