Integral of an Exponential function

rabbahs
Messages
16
Reaction score
0
Hello every one,

I was doing my research and then I simply struck at a point.
The point is that i do not know how to solve the following Integral. I am not at all bad at doing math but some times I got blanked.

so, here is the Integral,

Integral.jpg


Integral (infinity,u) exponent^(-u) du

result with derivation or with some reference will be highly appreciated.

thanks
 
Physics news on Phys.org
Err, that doesn't really make sense. Do you mean,
\int_{u}^\infty e^{-v} \, dv
by any chance (with the boundaries in the correct order and where the integration variable is a dummy not occurring in the integration boundary).
 
ok, if it is the case then what will be the answer ?
 
In a research paper, I found a solution to a similar problem.

Integral.jpg


please look at it
 
rabbahs said:
In a research paper, I found a solution to a similar problem.

View attachment 26661

please look at it
This is not that similar. Your problem, assuming that it is as CompuChip suggested, is
\int_u^{\infty} e^{-v}dv

First, find an antiderivative using substitution.
Second, evaluate the improper integral using limits.

This is not a very complicated integral.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top