Improper Integrals over Unbounded Interva

Improper integrals for infinite intervals are defined in a similar way

Suppose we want to calculate the integral of a continuous function over an unboundede interval of the from [a, +∞), (−∞, b] or (−∞, −∞). Like the intergral with unbounded integrand that we studied in the last section, these integrals are aid to be improper.

Definition 6.7.1. Let f : [a, +∞] ⟶ ℝ continuous. We define

$${\int }_a^{+\mathrm{\infty }}f(x)\text{d}x=\underset{\omega \to +\mathrm{\infty }}{lim}{\int }_a^{\omega }f(x)\text{d}x$$

If the limit exits finite then f is said integrable on [a, +∞) or that the integral is convergent. If the limit is ±∞ we say that the integral diverges. If, however, no such limit exists, we say that the improper integral does not exists.

An analogous definition is given if f: (−∞, b].

$${\int }_{-\mathrm{\infty }}^{b}f(x)\text{d}x=\underset{\omega \to -\mathrm{\infty }}{lim}{\int }_{\omega }^{b}f(x)\text{d}x$$

Finally if f: (−∞, ∞) ⟶ ℝ is continuos, we let

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f(x)\text{d}x={\int }_{-\mathrm{\infty }}^{c}f(x)\text{d}x+{\int }_c^{-\mathrm{\infty }}f(x)\text{d}x$$

for an arbitrary point c.

Example 6.7.2. Calculate the following imporper integral.

$${\int }_0^{\mathrm{\infty }}\frac{\text{d}x}{1+x^2}$$

we first calculate the integral

$${\int }_0^{\omega }\frac{\text{d}x}{1+x^2}$$

|tan⁻¹|₀^ω = tan⁻¹ ω − tan⁻¹ 0 = tan⁻¹ ω

The lim_{ω ⟶ ∞} tan⁻¹ ω = π/2. So the integral converges to π/2.  ■

Example 6.7.3. Evaluate the improper integral

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{1+x^2}\text{d}x$$

Solution. To evaluate this integral, we break the interval into two pieces (−∞, +∞) = (−∞, 0) ∪ [0, +∞). The choice of zero is arbitrary;

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{1+x^2}\text{d}x=\underset{\omega \to \mathrm{\infty }}{lim}\frac{1}{1+x^2}\text{d}x=\underset{\omega \to \mathrm{\infty }}{lim}{[\arctan (x)]}_0^{\omega }=\underset{\omega \to \mathrm{\infty }}{lim}(\arctan \omega -0)=\pi /2$$

The second integral evaluates to π/2 in a similar manner. We conclude that the original integral over the entire real line is convergent and that its value is π/2 + π/2 = π.  ■

Example 6.7.3. Evaluate the integral

$${\int }_1^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\text{d}x\alpha >0$$

Case α = 1. We have

$${\int }_1^{\omega }\frac{1}{x^{\alpha }}\text{d}x={[\log x]}_1^{\omega }=\log \omega$$

Since lim_{ω ⟶ +∞} log ω = +∞, the integral is divergent.

Case α ≠ 1. We have

$${\int }_1^{\omega }\frac{1}{x^{\alpha }}\text{d}x=\frac{1}{1-\alpha }{\left[x^{1-\alpha }\right]}_1^{\omega }=\frac{1}{1-\alpha }({\omega }^{1-\alpha }-1)$$

Thus

$${\int }_1^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\text{d}x=\underset{\omega \to +\mathrm{\infty }}{lim}\frac{1}{1-\alpha }({\omega }^{1-\alpha }-1)=\{\begin{array}{ll}+\mathrm{\infty }& \text{if }\alpha <1\\ \frac{1}{\alpha -1}& \text{if }\alpha >1\end{array}$$

Hence we have

$${\int }_1^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\text{d}x\text{is }\{\begin{array}{lll}& \text{divergent to }\mathrm{\infty }& \text{if }\alpha \le 1\\ & \text{convergent to }\frac{1}{\alpha -1}& \text{if }\alpha >1\end{array}$$

Divergence of the harmonic series

When we studied infinite series we stated that the harmonic series Σ 1/n is divergent; We shall now prove it. From figure 1, it can be easily seen that

$${\int }_1^N\frac{1}{x}\text{d}x<\sum _{n=1}^{N-1}\frac{1}{n}$$

Since ∫^N_{1 1/x dx = log N ⟶ +∞ if N ⟶ ∞, then ∑^Nn=1 1/n as well ⟶ +∞ if N ⟶ +∞, thus the harmonic series diverges.}

Congergence of the generalized harmonic series

We studied the generalized harmonic series

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{\alpha }}$$

converges for α > 1 (and diverges for α ≤ 1). We have proved for α ≥ 2 (by comparison with the Mengoli's series) and we are now going to provide a proof for any α > 1, in a similar manner used for the proof of conergence of the harmonic series. From fig. 2 it is clear that

$$\sum _{n=2}^N\frac{1}{n^{\alpha }}\le {\int }_1^N\frac{\text{d}x}{x^{\alpha }}$$

Since for α > 1 and N ⟶ ∞, the integral converges the series converges as well.

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