Calculate the improper integral: integral from -1 to 1

Calculate the improper integral: $$ \int_{-1}^1 \frac{dx}{\sqrt[3]{x}} $$

This is an improper integral because the integrand has a singularity at $x = 0$.

Step 1: Identify the singularity

The integrand is:

$$\frac{1}{\sqrt[3]{x}} = \frac{1}{x^{1/3}} = x^{-1/3}$$

At $x = 0$, this function is undefined and:

$$\lim_{x \to 0^+} x^{-1/3} = +\infty \quad \text{and} \quad \lim_{x \to 0^-} x^{-1/3} = -\infty$$

The singularity is at $x = 0$, which is in the interior of the interval $[-1, 1]$.

Step 2: Split the integral at the singularity

We must split the integral at $x = 0$:

$$\int_{-1}^1 \frac{dx}{\sqrt[3]{x}} = \int_{-1}^0 \frac{dx}{\sqrt[3]{x}} + \int_0^1 \frac{dx}{\sqrt[3]{x}}$$

For the integral to converge, both parts must converge.

Step 3: Analyze the first integral $\int_{-1}^0 \frac{dx}{\sqrt[3]{x}}$

For $x < 0$, we have $x = -|x|$, so:

$$\sqrt[3]{x} = \sqrt[3]{-|x|} = -\sqrt[3]{|x|} = -|x|^{1/3}$$

Therefore:

$$\frac{1}{\sqrt[3]{x}} = \frac{1}{-|x|^{1/3}} = -\frac{1}{|x|^{1/3}}$$

The integral becomes:

$$\int_{-1}^0 \frac{dx}{\sqrt[3]{x}} = \int_{-1}^0 -\frac{1}{|x|^{1/3}} \, dx$$

Let $u = -x$, so $x = -u$, $dx = -du$. When $x = -1$: $u = 1$; when $x = 0$: $u = 0$.

$$= \int_1^0 -\frac{1}{u^{1/3}} \cdot (-du) = -\int_1^0 \frac{du}{u^{1/3}} = \int_0^1 \frac{du}{u^{1/3}}$$

Step 4: Evaluate $\int_0^1 \frac{du}{u^{1/3}}$

This is of the form $\int_0^a \frac{dx}{x^\alpha}$ with $\alpha = \frac{1}{3}$.

Convergence criterion: This converges if and only if $\alpha < 1$.

Since $\frac{1}{3} < 1$, the integral converges! ✓

$$\int_0^1 u^{-1/3} \, du = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 u^{-1/3} \, du$$

Using the power rule:

$$\int u^{-1/3} \, du = \frac{u^{2/3}}{2/3} = \frac{3u^{2/3}}{2} + C$$

Evaluating:

$$= \lim_{\varepsilon \to 0^+} \left[\frac{3u^{2/3}}{2}\right]_\varepsilon^1 = \lim_{\varepsilon \to 0^+} \left(\frac{3(1)^{2/3}}{2} - \frac{3\varepsilon^{2/3}}{2}\right)$$

$$= \frac{3}{2} - 0 = \frac{3}{2}$$

Step 5: Evaluate the second integral $\int_0^1 \frac{dx}{\sqrt[3]{x}}$

This is exactly the same calculation:

$$\int_0^1 x^{-1/3} \, dx = \left[\frac{3x^{2/3}}{2}\right]_0^1 = \frac{3}{2} - 0 = \frac{3}{2}$$

Step 6: Combine the results

$$\int_{-1}^1 \frac{dx}{\sqrt[3]{x}} = \frac{3}{2} + \frac{3}{2} = 3$$

Wait, let me reconsider the sign more carefully...

Step 6 (Corrected): Careful analysis of signs

For $x \in [-1, 0)$: $x < 0$, so $x^{-1/3} = \frac{1}{x^{1/3}}$

The cube root of a negative number is negative: $(-1)^{1/3} = -1$

More generally, for $x = -a$ where $a > 0$:

$$x^{-1/3} = (-a)^{-1/3} = \frac{1}{(-a)^{1/3}} = \frac{1}{-a^{1/3}} = -\frac{1}{a^{1/3}} = -a^{-1/3}$$

So the integrand is negative for $x \in [-1, 0)$ and positive for $x \in (0, 1]$.

Let's recalculate:

$$\int_{-1}^0 x^{-1/3} \, dx = \lim_{\varepsilon \to 0^-} \int_{-1}^\varepsilon x^{-1/3} \, dx$$

$$= \lim_{\varepsilon \to 0^-} \left[\frac{3x^{2/3}}{2}\right]_{-1}^\varepsilon$$

For $x = \varepsilon < 0$: $x^{2/3} = (x^2)^{1/3} = |\varepsilon|^{2/3} = |\varepsilon|^{2/3}$ (always positive)

For $x = -1$: $(-1)^{2/3} = [(-1)^2]^{1/3} = 1^{1/3} = 1$

$$= \lim_{\varepsilon \to 0^-} \left(\frac{3|\varepsilon|^{2/3}}{2} - \frac{3(1)}{2}\right) = 0 - \frac{3}{2} = -\frac{3}{2}$$

For the second integral:

$$\int_0^1 x^{-1/3} \, dx = \left[\frac{3x^{2/3}}{2}\right]_0^1 = \frac{3}{2} - 0 = \frac{3}{2}$$

Step 7: Final combination

$$\int_{-1}^1 \frac{dx}{\sqrt[3]{x}} = -\frac{3}{2} + \frac{3}{2} = 0$$

Final Answer

$$\boxed{\int_{-1}^1 \frac{dx}{\sqrt[3]{x}} = 0}$$

The integral converges to 0.

Explanation of the result

The result is 0 because the function $f(x) = x^{-1/3}$ is an odd function:

$$f(-x) = (-x)^{-1/3} = -x^{-1/3} = -f(x)$$

For odd functions integrated over symmetric intervals $[-a, a]$:

$$\int_{-a}^a f(x) \, dx = 0$$

This is because the negative area on $[-a, 0]$ exactly cancels the positive area on $[0, a]$.

Verification of odd function property

Let's verify that $x^{-1/3}$ is odd:

$$(-x)^{-1/3} = \frac{1}{(-x)^{1/3}} = \frac{1}{-(x^{1/3})} = -\frac{1}{x^{1/3}} = -x^{-1/3}$$

Yes, $f(-x) = -f(x)$, so it's an odd function. ✓

Geometric interpretation

The graph of $y = x^{-1/3} = \frac{1}{\sqrt[3]{x}}$ has these properties:

• Vertical asymptote at $x = 0$ (goes to $+\infty$ from the right, $-\infty$ from the left)
• Symmetric about the origin: if $(a, b)$ is on the graph, so is $(-a, -b)$
• For $x > 0$: the function is positive and decreasing
• For $x < 0$: the function is negative and increasing

The signed areas on $[-1, 0]$ and $[0, 1]$ are equal in magnitude but opposite in sign, so they cancel.

Why both integrals converge

Both $\int_{-1}^0 \frac{dx}{\sqrt[3]{x}}$ and $\int_0^1 \frac{dx}{\sqrt[3]{x}}$ have the form $\int \frac{dx}{x^\alpha}$ with $\alpha = \frac{1}{3} < 1$.

Since $\alpha < 1$, both integrals converge individually:

• $\int_{-1}^0 \frac{dx}{\sqrt[3]{x}} = -\frac{3}{2}$ (finite, negative)
• $\int_0^1 \frac{dx}{\sqrt[3]{x}} = \frac{3}{2}$ (finite, positive)

Their sum is 0.

Comparison with other exponents

Case 1: $\int_{-1}^1 \frac{dx}{\sqrt{x}} = \int_{-1}^1 x^{-1/2} \, dx$

This integral has problems because $x^{-1/2}$ is not real for $x < 0$ (in the real numbers). If we consider only $[0, 1]$, it converges since $\frac{1}{2} < 1$.

Case 2: $\int_{-1}^1 \frac{dx}{\sqrt[3]{x}} = \int_{-1}^1 x^{-1/3} \, dx = 0$

This converges because $\frac{1}{3} < 1$ and the function is odd.

Case 3: $\int_{-1}^1 \frac{dx}{x} = \int_{-1}^1 x^{-1} \, dx$

This diverges because $\alpha = 1$, even though the function is odd. The Cauchy principal value exists and equals 0, but the improper integral diverges.

Summary

1. The integral has a singularity at $x = 0$ (interior point)
2. We split: $\int_{-1}^1 = \int_{-1}^0 + \int_0^1$
3. The exponent $\alpha = \frac{1}{3} < 1$ ensures both parts converge
4. The function $x^{-1/3}$ is odd: $f(-x) = -f(x)$
5. For odd functions on symmetric intervals: $\int_{-a}^a f(x) \, dx = 0$
6. Calculation gives: $-\frac{3}{2} + \frac{3}{2} = 0$

Key lessons:

• Interior singularities require splitting the integral
• Odd functions integrated over symmetric intervals give 0
• The convergence criterion $\alpha < 1$ applies at each singularity
• Cube roots (and odd roots) preserve signs: $\sqrt[3]{-8} = -2$
• Both parts must converge for the total integral to converge
• Symmetry can greatly simplify calculations and provide insight