Solving Exponential Integral with Pathlengths for Solids

In summary, the conversation discusses attempting to solve an integral involving exponential and square root functions. The individual tried using substitution and integration by parts methods but was unsure if their solution was correct. They then asked for confirmation on their method and solution.
  • #1
k_c
2
0

Homework Statement



I'm working out different pathlengths for different solids but I'm stuck on the following integral:

Homework Equations



[tex]\int[/tex] x*exp(-C*sqrt[1-x2/A2]) dx

exp = exponential func
sqrt = square root
C and A are constants

The Attempt at a Solution



I tried to work it out with the substitution method, where u = -C*sqrt[1-x2/A2] and du = C/A2x(1-x2/A2)-1/2.

But it seems to be getting really complex afterwards, so I was wondering if I'm overlooking the simple approach for this integral?
 
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  • #2
try integration by parts.
 
  • #3
I tried it this way now:

u^2 = (1-x^2/a^2)
so du = -2x/a^2 dx and therefore x dx = -a^2/2 du

So the integral becomes: [tex]\int[/tex] -a^2/2 exp^(-cu) du
And then i find the solution: a^2/(2c) exp^(-cu)

Which is a^2/(2c) exp^[-c sqrt(1-x^2/a^2)]


I think this should be correct but the wolfram mathematica integrator gives me a different solution, so if somebody could confirm my method/solution, that would be great..
 
  • #4
k_c said:
I tried it this way now:

u^2 = (1-x^2/a^2)
so du = -2x/a^2 dx and therefore x dx = -a^2/2 du

Check du.

ehild
 
  • #5
For the integral:
[tex]
\int x\exp (-c\sqrt{1-x^{2}/a^{2}})dx
[/tex]
I use the substitute:
[tex]
u^{2}=1-\frac{x^{2}}{a^{2}}
[/tex]
Then:
[tex]
xdx=a^{2}udu
[/tex]
This makes the integral become:
[tex]
a^{2}\int ue^{-u}du
[/tex]
This I think is easy to compute, so I will leave that part to you.
 

FAQ: Solving Exponential Integral with Pathlengths for Solids

What is an exponential integral?

An exponential integral is a mathematical function that represents a generalization of the logarithm function. It is often used to solve problems involving exponential growth or decay.

How is the exponential integral used in solving problems involving pathlengths for solids?

The exponential integral is used to calculate the pathlengths for solids by considering the absorption and scattering of light within the material. The integral takes into account the optical properties of the material, such as its refractive index and absorption coefficient, to determine the pathlength of light through the solid.

What is the significance of pathlengths for solids?

Pathlengths for solids are important in understanding how light interacts with materials. They are used in a variety of applications, including in the design of optical devices and in the study of light-matter interactions in materials.

Can the exponential integral be solved analytically?

Yes, the exponential integral can be solved analytically using various mathematical techniques, such as integration by parts or substitution. However, for more complex problems, numerical methods may be required to obtain a solution.

What are some limitations of using the exponential integral to solve problems involving pathlengths for solids?

One limitation is that the exponential integral assumes a homogeneous and isotropic medium, which may not always be the case in real-world materials. Additionally, the integral may become more complex and difficult to solve for materials with non-linear optical properties.

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