0 to positive infinity e xdx

(0,+∞e^-xdx=1。

The process is as follows:

e^(-x)dx

e^(-x)d(-x)

e (-x) +c, where c is constant.

So. (0,+∞e^(-x)dx

e (-x) to substitute the upper and lower limits + and 0

e^(-e^0

Apparently e (-=0, and e 0=1

So. (0,+∞e^(-x)dx

e^(-e^0

(0 to positive infinity) e -xdx=- (0 to positive infinity) e -xd(-x)=lim(x--> e (-x)-lim(x-->0)e -x=-1

Answer: 0 )e (-2x)dx

0 and bend) (1 high shed oak next to 2) e (-2x) d (-2x).

0→∞)1/2)e^(-2x)

<> here we take the real part of the calculation, so the final result only needs to take the real part.

The latter integral is an imaginary number.

It can be omitted after taking the real part.

If the sequence converges, it must be bounded. i.e. for everything n(n=1,2......You can always find a positive number m, so that |xn|≤m。

(0 to positive scum infinity) e -xdx=- e.g. renting (0 to positive infinity) e -xd(-x)=lim(x-->e (-x)-lim(x--> imitation 0)e -x=-1

Use the gamma function.

It is more convenient with the residual element formula.

x) = t (x-1) e t dt integral limit is 0 to positive infinity.

Take x=3 2.

1 2)= t (-1 2) *e (-t)dt = 1 x * e (-x 2) d(x 2)=2 picos e (-x 2)dx

The remainder formula is.

x)*γ1-x)=πsinπx

So (1 2) =

So. e (-x 2)dx = 1 2) 2 = 2 Another way is to calculate.

The value of e(-x 2+y 2))dxdy on [0,r][0,r], this calculation is first converted to the pole seat.

The pinch principle is then used to find the limit.

Then open the square.

(xe^x)(-0]-∫0]e^xdx

xe^x)(-0]-e^x(-∞0]

Let's prove that Yuanxin (xe x) (-0] 0

lim(x→－∞xe^x

lim(x→－∞x/e^(-x) (

lim(x,