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For $p,q>0$, express the product of the two Gamma integrals as the nonnegative double integralUse $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and Jacobian determinant of absolute value $r$. The new domain of integration is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}{e}^{-r}{u}^{p-1}(1-u{)}^{q-1}$. Tonelli's theorem and separation of the two integrals therefore give $\Gamma (p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.

For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma (p+q)B(p,q)=\Gamma (p)\Gamma (q)$. Absolute integrability follows by taking real parts in the exponents, so Fubini's theorem applies. The nonvanishing of the Gamma function then permits division by $\Gamma (p+q)$.

For integer $n\ge 2$, integration by parts gives $J_n=(n-1)J_{n-2}/2$. The boundary term $-x^{n-1}{e}^{-x^2}/2$ vanishes at both infinities because Gaussian decay dominates every power. The full-line initial values are $J_0=\sqrt{\pi }$ and $J_1=0$, the latter by odd symmetry. HenceUnlike the half-line Gaussian moment recurrence, full-line odd moments vanish and even moments are twice their half-line values. The domain of integration must be retained when transferring a recurrence between the two settings.

An integral adds up contributions of a function across a domain of integration. For a continuous function $f$ on $[a,b]$, the Riemann integral is the common limit of its Riemann sums as the largest subinterval width tends to zero:Each product combines a value with the width over which it acts. When $f$ is nonnegative, the result is the area beneath its graph; values below the horizontal axis contribute with the opposite sign. The signed area under a graph can therefore vanish even when the graph encloses nonzero geometric area.

For a velocity $v(t)=1+2t$ on $0\le t\le 3$, use $n$ equal time steps. The left and right Riemann sums areBecause the function is increasing, they bound the integral. With six steps they give $10.5$ and $13.5$; with sixty steps they give $11.85$ and $12.15$. Both approach the same total, $12$. This is accumulation made exact by refining finite measurements.

An integral retains units: velocity times time gives displacement, and a probability density function integrated over an interval gives a probability. Weighting a density by a function gives its expectation. The same operation serves different subjects because it expresses a common pattern, value multiplied by extent, then accumulated.

The fundamental theorem of calculus often evaluates a definite integral by finding an antiderivative, but a finite sum or a numerical approximation is still useful when no convenient antiderivative is available. An indefinite integral denotes a family of antiderivatives; a definite integral denotes an accumulated value between specified endpoints. More general constructions, including Lebesgue integration, extend the class of functions and domains that can be handled.

The sequence in which variables are integrated in an iterated integral. Reversing the order changes the slicing of the domain of integration, often requiring more than one integral. Fubini's theorem permits interchange for absolutely integrable functions, while Tonelli's theorem permits nonnegative measurable integrands even when the integral is infinite. Conditional convergence requires additional care; geometric reversal alone is not a general justification for interchanging improper integrals.

The domain of integration is $0\le x\le 2$, $0\le y\le x^3$. Reversing the order in the double integral and then putting $u=x^4+1$ givesTherefore

The domain of integration is the upper half of the ellipse $x^2+4y^2\le 1$. Reversing the order makes $-1\le x\le 1$ and $0\le y\le \frac{1}{2}\sqrt{1-x^2}$. ThusThe inner antiderivative is $y^2/2$, so the square of the upper limit removes the square root.