Pls help me to integrate Exponent^(-0.5 x^2) with respect to dx?

Question

Exponential Integration

Answer 1

The integral of e^(x^2) has no closed formula.
Check out this link:
http://en.wikipedia.org/wiki/Error_function

Answer 2

Try integration by parts

⌠u∙dv = uv - ⌠v∙du

It doesn't look as if that will ever give you a 'do-able' integral, but the intermediaries terms (the uv terms) generated at each repeated application may form a power series expansion of some function that has
e^(-.5x^2) as its argument.

If that goes nowhere, go back to the series expansion of e^x

e^x = Σ (x^n) / (n!) for r = 0 to r = ∞

Integrate the first dozen or so terms and look for any 'patterns' or ways to combine terms to get into some standard form expansion of a common (or even not so common ) function.

This is where a good table of series and integrals comes in *real* handy.

It's probably going to take several hours of 'cut and try' to solve this one. If it's any consolation, I've had problems that took several *months* to get into a canonical form solution

Good luck.


Doug

Answer 3

you could obtain a simple solution by expanding the exponential as in: exp(-0.5x^2) = 1-0.5x^2+(0.5x^2)^2/2!-+......
if you assume x to be small, only the first 2 terms of the expansion are a sufficient approximation.
thus, your integral will give the result: x - x^3/6

Answer 4

A closed function, it doesn't exist, but we can find its value using polar coordinates while x is defined from - ∞ to +∞

∫℮^(-x²/2) dx where x takes values from - ∞ to +∞

Let say that we want to evaluate this
∫∫℮^[-(x²+y²)/2] dxdy

Using polar coordinates is: r ²=x²+y²
x=rcosθ, y = rsinθ
and dx dy = dA using the Jacobian:
dA = rdrdθ
So,
∫∫℮^(-r²/2) rdrdθ Now, our limits will be
r=0
r=∞
θ=0
θ=2π

Then now we can integrate respect to dθ first, where θ is defined from 0 to 2π:

∫∫℮^(-r²/2) rdrdθ = ∫θ℮^(-r²/2) rdrθ = 2π∫℮^(-r²/2) r dr

Let u = r²/2 du = rdr

2π∫℮^(-u) du = 2π [-℮^(-u)] = -2π℮^(-r²/2)

= 2π[℮^(-0²/2) - lim r →∞ ℮^(-r²/2)] = 2π[1 - 0] = 2π

but as
∫℮^(-x²/2) dx= ∫℮^(-y²/2)dy and
∫∫℮^[-(x²+y²)/2]dxdy =[ ∫℮^(-x²/2) ] ²

Then
∫∫℮^(-r²/2) rdrdθ = [ ∫℮^(-x²/2)dx ] ² = 2π

Then
∫℮^(-x²/2) dx = √(2π) <<<<<<<<<<<<<<<<<<<<<<

where x is defined from - ∞ to +∞

Answer 5

There is no closed-form solution for this. However, that integral is equivalent to

Exponent(-0.5 x^2) = 1.25331*erf( x / sqrt(2) )

where erf is the error function (see http://mathworld.wolfram.com/Erf.html)

I hope this helps.

Answer 6

you can try this question by integrating by parts

Answer 7

if i understand ur question
f(x) = e^(-0.5x^2)
f'(x) = [e^ (-0.5x^2)][-x]
f'(x) = -xe^(-0.5x^2)