Search results
Results from the WOW.Com Content Network
The following is a list of integrals (antiderivative functions) of logarithmic functions. For a complete list of integral functions, see list of integrals. Note: x > 0 is assumed throughout this article, and the constant of integration is omitted for simplicity.
The logarithmic integral has an integral representation defined for all positive real numbers x ≠ 1 by the definite integral li ( x ) = ∫ 0 x d t ln t . {\displaystyle \operatorname {li} (x)=\int _{0}^{x}{\frac {dt}{\ln t}}.}
In mathematics, the definite integral ∫ a b f ( x ) d x {\displaystyle \int _{a}^{b}f(x)\,dx} is the area of the region in the xy -plane bounded by the graph of f , the x -axis, and the lines x = a and x = b , such that area above the x -axis adds to the total, and that below the x -axis subtracts from the total.
If the function f does not have any continuous antiderivative which takes the value zero at the zeros of f (this is the case for the sine and the cosine functions), then sgn(f(x)) ∫ f(x) dx is an antiderivative of f on every interval on which f is not zero, but may be discontinuous at the points where f(x) = 0.
ln(r) is the standard natural logarithm of the real number r. Arg(z) is the principal value of the arg function; its value is restricted to (−π, π]. It can be computed using Arg(x + iy) = atan2(y, x). Log(z) is the principal value of the complex logarithm function and has imaginary part in the range (−π, π].
The function f(x) is called the integrand, the points a and b are called the limits (or bounds) of integration, and the integral is said to be over the interval [a, b], called the interval of integration. [18] A function is said to be integrable if its integral over its domain is finite. If limits are specified, the integral is called a ...
Integrands of the form x m (a + b x n + c x 2n) p when b 2 − 4 a c = 0 [ edit ] The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.