# Integration of an exponential function

• Safinaz
In summary, integrating the function ##e^{-2\alpha(y)}## is not as simple as just taking the negative of the same function. The difficulty lies in the fact that ##\alpha(y)## is a function of y, and its specific form must be known in order to properly integrate the function. If ##\alpha(y)## is a linear function of y, the integration is straightforward, but for more complicated functions, numerical methods may be required. There may be special cases where an analytical solution exists, but in general, the integral cannot be solved without knowing the form of ##\alpha(y)##.

#### Safinaz

Homework Statement
Hello ,

How to integrate
Relevant Equations
## \int ~ dy ~ e^{-2 \alpha(y)} ##
My trial :

I think ## \int ~ dy ~ e^{-2 \alpha(y)} ## dose not simply equal: ## - \frac{1}{2}e^{-2 \alpha(y)} ## cause ##\alpha## is a function in ##y ##.

Delta2
What is ##\alpha(y)## ?

anuttarasammyak said:
What is ##\alpha(y)## ?
Should I assume it to make the integration, right?
Well, in this case let's assume it increases with y exponentially or it's a slowly varying function

Safinaz said:
Should I assume it to make the integration, right?
Well, in this case let's assume it increases with y exponentially or it's a slowly varying function
You need to specify the function ##\alpha(y)## - i.e. give the actual formula for ##\alpha## in terms of y. Or (if integrating numerically) you need a table giving values of ##\alpha## for values of y over the range of interest.

If ##\alpha(y)## can be represented as a linear function of y (##\alpha(y) = ay + b## with a and b as constants) then the integration is clearly simple.

For more complicated functions, I believe there are no general analytical methods, though special cases may have solutions . E.g. with ##\alpha(y) = y^2## the integral can be expressed in terms of the error function – see https://www.wolframalpha.com/input/?i=e^(-2y^2)

Also, see discussion here: https://math.stackexchange.com/questions/19390/integrating-efx

Omega0, docnet, Safinaz and 2 others
@Steve4Physics. Hay! just saying thank you very much! The answer is so helpful

Delta2 and Steve4Physics