Improper integrals
Problem 4.36 · easy
Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x} \, dx \), or show that it diverges.
- The integral is improper at infinity; replace that bound by T and take a limit.
- \[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]An antiderivative, checked by differentiating.✓ Proved
- \[ \ln{\left(T \right)} \]The integral with T in place of the bad bound.✓ Proved
- As T → \infty, this grows without bound, so the integral diverges.
Answer \( \text{diverges} \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical integrals to 10², 10⁴, 10⁶ keep growing |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/improper_integral, checked 2026-09-26 with SymPy 1.14.0.