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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. Integrals involving only logarithmic functions
Plot of the logarithmic integral function li(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D. In mathematics, the logarithmic integral function or integral logarithm li(x) is a special function. It is relevant in problems of physics and has number theoretic significance.
This can be proved by computing the derivative of the right-hand side of the formula, taking into account that the condition on g is here for insuring the continuity of the integral. This gives the following formulas (where a ≠ 0), which are valid over any interval where f is continuous (over larger intervals, the constant C must be replaced ...
(Note that the value of the expression is independent of the value of n, which is why it does not appear in the integral.) ∫ x x ⋅ ⋅ x ⏟ m d x = ∑ n = 0 m ( − 1 ) n ( n + 1 ) n − 1 n !
As an integral, ln(t) equals the area between the x-axis and the graph of the function 1/x, ranging from x = 1 to x = t. This is a consequence of the fundamental theorem of calculus and the fact that the derivative of ln(x) is 1/x. Product and power logarithm formulas can be derived from this definition. [41]
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 natural logarithm of x is generally written as ln x, log e x, or sometimes, if the base e is implicit, simply log x. [2] [3] Parentheses are sometimes added for clarity, giving ln(x), log e (x), or log(x). This is done particularly when the argument to the logarithm is not a single symbol, so as to prevent ambiguity. The natural logarithm ...
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.