I can see two different ways to place the parentheses in that question.
Here are both answers:
( e-2 ) x infinity = infinity
( e-2 x infinity ) = zero
E to the power infinity, or lim en as n approaches infinity is infinity.
that would be the inverse of e to the plus infinity Answer is thus zero
Trans infinity
This browser is not much use when it comes to mathematics but I'll try.Suppose X is a random variable with a Normal distribution and let f(x) be the probability density function of x.Then the mean is mu = E(X) = Integral of x*f(x) dx over the domain of X [which is negative infinity to positive infinity].The variance is E{[X - E(X)]2} = Integral of (x - mu)2*f(x) dx over the domain of X.
???
As x tends to negative infinity, the expression is asymptotically 0.
10 (or e) to the power of x range from zero to infinity. Lets try the extreme cases: 10^infinity = infinity 10^0 = 1 10^-infinity = 1/infinity = 0
Int (e^(-x^2)) = Int (1 + (-x^2) + (-x^2)^2/2! + (-x^2)^3/3! + ... = x - x^3/3 + x^5/(5*2!) - x^7/(7*3!) ... which, if taken with limits of integration from negative infinity to infinity, solves to the square root of x, making it one of the most famous and beautiful formulas in math.
E to the power infinity, or lim en as n approaches infinity is infinity.
6 E -2 (six times ten to the power of negative-two)
infinity
Infinity
that would be the inverse of e to the plus infinity Answer is thus zero
Trans infinity
$$CLPBRD$$e@Sup{-2} or constant e to the -2(negative squared!)
sum from{-infinity } to{infinity } ({1} over {2 * pi } )(int (abs{X(func e^{jw})})^2 )dw
This browser is not much use when it comes to mathematics but I'll try.Suppose X is a random variable with a Normal distribution and let f(x) be the probability density function of x.Then the mean is mu = E(X) = Integral of x*f(x) dx over the domain of X [which is negative infinity to positive infinity].The variance is E{[X - E(X)]2} = Integral of (x - mu)2*f(x) dx over the domain of X.