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The Improper Integral Calculator determines whether an integral with an infinite upper limit converges or diverges, and computes the exact value when convergent. It handles two important families of functions: power-type functions $$\frac{1}{x^p}$$ and exponential decay functions $$e^{-ax}$$.
An improper integral is a definite integral where either the interval of integration is unbounded (extends to infinity) or the integrand has a singularity within the interval. This calculator focuses on the first type: integrals of the form $$\int_c^{\infty} f(x)\,dx$$, where the upper limit is infinity. Such integrals arise naturally when computing total quantities over unlimited domains.
The key question for any improper integral is convergence: does the integral have a finite value, or does it grow without bound? The answer depends critically on how fast the function decays as $$x \to \infty$$. If the function decays rapidly enough, the accumulated area remains finite despite the infinite interval.
For the p-integral $$\int_c^{\infty} \frac{1}{x^p}\,dx$$, convergence depends entirely on $$p$$. When $$p > 1$$, the integral converges to $$\frac{c^{1-p}}{p-1}$$. When $$p \leq 1$$, the integral diverges. The critical case $$p = 1$$ gives $$\int_1^{\infty} \frac{1}{x}\,dx = \ln(\infty) = \infty$$, the famous harmonic divergence. This sharp threshold at $$p = 1$$ is a foundational result in analysis.
For exponential decay $$\int_c^{\infty} e^{-ax}\,dx$$ with $$a > 0$$, the integral always converges because exponential decay dominates any polynomial growth. The value is $$\frac{e^{-ac}}{a}$$. Exponential integrals appear in probability (expected values of exponential distributions), radioactive decay, heat transfer, and signal processing.
Select your function type, enter the parameters, and the calculator will instantly determine convergence, compute the value if convergent, and display the convergence condition and antiderivative form.
The calculator evaluates improper integrals using limit definitions and closed-form antiderivatives.
Power type: $$\int_c^{\infty} \frac{1}{x^p}\,dx$$
The antiderivative of $$x^{-p}$$ for $$p \neq 1$$ is $$\frac{x^{1-p}}{1-p}$$. Evaluating at the bounds:
$$\int_c^{\infty} x^{-p}\,dx = \lim_{t \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_c^t = \lim_{t \to \infty} \frac{t^{1-p}}{1-p} - \frac{c^{1-p}}{1-p}$$
When $$p > 1$$, the exponent $$1 - p < 0$$, so $$t^{1-p} \to 0$$ as $$t \to \infty$$. The integral converges to $$\frac{c^{1-p}}{p - 1}$$. When $$p \leq 1$$, $$t^{1-p} \to \infty$$, so the integral diverges.
Exponential type: $$\int_c^{\infty} e^{-ax}\,dx$$
The antiderivative is $$-\frac{1}{a}e^{-ax}$$. As $$t \to \infty$$, $$e^{-at} \to 0$$:
$$\int_c^{\infty} e^{-ax}\,dx = 0 - \left(-\frac{1}{a}e^{-ac}\right) = \frac{e^{-ac}}{a}$$
Convergence tells you whether the integral has a finite value. "Convergent" means the total area under the curve from the lower bound to infinity is finite. "Divergent" means the area grows without bound.
The Integral Value is displayed only when the integral converges. For divergent integrals, the value is shown as 0 (not meaningful). The value represents the total area under the curve from the lower bound extending infinitely to the right.
The Convergence Condition explains the mathematical criterion. For power functions, $$p > 1$$ is required. For exponentials, any positive rate $$a > 0$$ suffices.
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With p = 2 and lower bound 1: value = 1^(1−2)/(2−1) = 1/1 = 1. This is the classic result that the area under 1/x² from 1 to infinity equals exactly 1.
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Value = e^(−2·0.5)/2 = e^(−1)/2 = 0.3679/2 = 0.1839. Exponential decay always produces a finite integral for positive rate parameters.
An integral is improper if the interval of integration is infinite (e.g., $$\int_1^{\infty}$$) or if the integrand has a discontinuity or singularity within the interval. This calculator handles the first type — integrals over infinite intervals.
The function $$1/x$$ decays too slowly: its antiderivative $$\ln x$$ grows without bound. Meanwhile $$1/x^2$$ decays fast enough that its antiderivative $$-1/x$$ approaches a finite limit. The critical threshold is $$p = 1$$: convergence requires $$p > 1$$.
The p-test states that $$\int_c^{\infty} \frac{1}{x^p}\,dx$$ converges if and only if $$p > 1$$ (assuming $$c > 0$$). When $$p = 1$$, the integral equals $$\ln(\infty) - \ln(c) = \infty$$. When $$p < 1$$, it diverges even faster.
Exponential decay $$e^{-ax}$$ with $$a > 0$$ decreases faster than any power of $$x$$. The function approaches zero so rapidly that the total area under the curve from any finite starting point to infinity remains bounded.
Yes, if the integrand is negative over the integration interval. For the functions in this calculator ($$1/x^p$$ with $$x > 0$$ and $$e^{-ax}$$), values are always positive, so the integrals are positive. In general, improper integrals can be negative or zero.
Many probability distributions are defined over infinite intervals. For example, the exponential distribution has PDF $$f(x) = ae^{-ax}$$ for $$x \geq 0$$, and its total probability is $$\int_0^{\infty} ae^{-ax}\,dx = 1$$. Expected values and moments also involve improper integrals.
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