A converging tail
- Input
- ∫ de 1 a inf de 1/x^2 dx
- Expected output
- 1 (converge)
p = 2 > 1.
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improper integral calculator
Type inf or -inf as one of the limits, or choose an interval where the function blows up. The calculator locates the singularities, probes the limit of the antiderivative at each problem point and answers converges or diverges, with the value when it exists.
p = 2 > 1.
F = −e⁻ˣ.
F = arctan x, from −π/2 to π/2.
x·ln x − x tends to 0 at 0.
Singularity at x = 0.
The indefinite integral of f is the family of functions F whose derivative is f. Since the derivative of a constant is zero, if F is an antiderivative then F + C is too, for any number C. That is why ∫ 2x dx = x² + C: x² + 1 and x² − 7 also have derivative 2x. The constant can only be pinned down with an extra condition, such as F(0) = 3.
It probes the antiderivative near the problem point at decreasing scales. If the values settle, the limit exists; if they grow or oscillate, it does not. An independent numerical quadrature serves as a second opinion.
No. In the integral from −1 to 1 of 1/x dx the principal value would be 0 by symmetry, but the improper integral in the usual sense diverges, and that is what the tool reports.
Yes: -inf and inf. The integral is split and each half must converge. That is why the integral from −∞ to ∞ of x dx diverges, even though it looks like 0 by symmetry.
Differentiating the result
F′(x) = f(x) at 14 test points (maximum error 2.2e-16). Both expressions are also identical after expanding.
| x | f(x) | F′(x) | Difference |
|---|---|---|---|
| -2.26376 | -0.23533751 | -0.23533751 | 2.8e-17 |
| 0.565523 | 0.99552794 | 0.99552794 | 1.1e-16 |
| 4.44149 | 377.08727 | 377.08727 | 5.7e-14 |
| -0.222645 | -0.17820488 | -0.17820488 | 0 |
| -0.357476 | -0.25003303 | -0.25003303 | 0 |
| 0.460526 | 0.72989202 | 0.72989202 | 2.2e-16 |
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