Improper Integrals for JEE
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Direct answer
When infinity enters an integral — as an infinite limit or as a blow-up at a finite point — the definition replaces the offending endpoint with a variable and takes a limit; a finite limit means the integral converges, otherwise it diverges. The p-test is the whole subject in two lines: ∫₁^∞ dx/x^p converges exactly when p > 1, while ∫₀¹ dx/x^p converges exactly when p < 1 — the same integrand, opposite verdicts, because the danger moves from infinity to the origin. Standard values worth carrying: ∫₀^∞ e^(−x) dx = 1, ∫₀^∞ dx/(1 + x²) = π/2, ∫₀^∞ e^(−x²) dx = √π/2, and ∫₀¹ ln x dx = −1. Interior singularities demand splitting — both halves must converge.
What you must remember
- Definition: ∫ₐ^∞ f(x) dx = lim b→∞ ∫ₐ^b f(x) dx, and similarly at a finite singularity approached from the safe side.
- p-test at infinity: ∫₁^∞ x^(−p) converges iff p > 1 — 1/x² converges, 1/x and 1/√x diverge.
- p-test at the origin: ∫₀¹ x^(−p) converges iff p < 1 — the verdicts flip because the danger relocated.
- Standard values: ∫₀^∞ e^(−x) dx = 1; ∫₀^∞ dx/(1 + x²) = π/2; ∫₀¹ ln x dx = −1; ∫₀^∞ dx/(x² + a²) = π/(2a).
- The Gaussian: ∫_{−∞}^{∞} e^(−x²) dx = √π, the bell-curve area that no elementary antiderivative ever produces.
- Splitting rule: for a singularity at an interior point c, write ∫ₐ^c + ∫c^b and demand convergence of each — one divergent half condemns the whole.
- Gamma connection: Γ(n) = ∫₀^∞ x^(n−1) e^(−x) dx with Γ(n) = (n − 1)! and Γ(1/2) = √π, the doorway Advanced papers occasionally open.
Two convergent evaluations
First, ∫₀^∞ dx/(x² + 4). The antiderivative is (1/2) arctan(x/2), so the integral is lim b→∞ (1/2) arctan(b/2) − 0 = (1/2)(π/2) = π/4 — convergence is visible as arctan squeezing toward its ceiling. Second, ∫₀¹ ln x dx. The antiderivative x ln x − x is undefined at 0 in the naive sense, so take the limit: as x → 0⁺, x ln x → 0 (exponentials beat logarithms), leaving (1 × 0 − 1) − 0 = −1. Both examples show the same discipline: compute the antiderivative, then let the endpoint run, evaluating the limit rather than substituting blindly. Now the cautionary third: ∫₀^∞ dx/(x(x + 1)) looks harmless, and at infinity it behaves like 1/x², converging there; but near 0 it behaves like 1/x, and ∫₀¹ dx/x diverges — the whole integral diverges from the origin side. Convergence is a local question asked at every suspicious point, not just at infinity.
Divergence hiding in plain sight
JEE Main asks the direct evaluations — arctan and exponential types — and the p-test as a convergent-or-not classification, where the distractors are always the flipped verdicts from testing 1/x^p at the wrong end. JEE Advanced enjoys disguised singularities: the x(x + 1) example above, or ∫₋₁¹ dx/x², which the careless evaluate as −2 after substituting, though the integral diverges at the interior point 0 — the positive integrand cannot integrate to a negative number, and that contradiction is the tell. Gamma and beta recognition appears occasionally: ∫₀^∞ x³ e^(−x) dx is Γ(4) = 3! = 6, computable by repeated integration by parts or by recognition in one glance. The splitting rule is where marks systematically die — candidates integrate across an interior asymptote using a single antiderivative and report a finite number for a divergent integral. The habit that saves: before any computation, scan the interval for zeros of denominators and infinities of integrands, and split first.
Frequently asked questions
When does ∫₁^∞ dx/x^p converge?
Exactly when p > 1; at p = 1 the integral of 1/x grows like ln b and diverges logarithmically.
Why does the verdict flip on [0, 1]?
There the danger is the origin, not infinity: ∫₀¹ dx/x^p converges exactly when p < 1.
What is ∫₀^∞ e^(−x²) dx?
√π/2 for the half-line, √π for the whole line — the Gaussian values with no elementary antiderivative behind them.
Why must you split at an interior singularity?
Convergence of ∫ₐ^b with a blow-up at c means both ∫ₐ^c and ∫c^b converge separately; cross-cancellation between divergent halves is forbidden by definition.
How does the gamma function connect?
Γ(n) = ∫₀^∞ x^(n−1) e^(−x) dx equals (n − 1)! for whole n and √π at n = 1/2, converting integrals into factorials.