Integrate exp(-(x^2)) using the substitution u=tanh(x)

In summary, the question is to convert the infinity limits of the integral \int^\infty_{-\infty} e^{{-x}^2} dx to finite limits \int^{u_a}_{u_b} g(u) du using the substitution u=tanh(x). To do this, the user uses the hint that tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x). They then use the substitution u=tanh(x) to solve for g(u). However, the user runs into trouble when they try to simplify the result of the integral by using the derivative. They end up with \
  • #1
sai2020
26
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The question is to convert the infinity limits of the integral [tex]\int^\infty_{-\infty} e^{{-x}^2} dx[/tex] to finite limits [tex]\int^{u_a}_{u_b} g(u) du[/tex] using the substitution [tex]u = tanh(x)[/tex].

How do I go about it?
 
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  • #2
sai2020 said:
The question is to convert the infinity limits of the integral [tex]\int^\infty_{-\infty} e^{{-x}^2} dx[/tex] to finite limits [tex]\int^{u_a}_{u_b} g(u) du[/tex] using the substitution [tex]u = tanh(x)[/tex].

How do I go about it?

Hi sai2020! :smile:

Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) :smile:
 
  • #3
tiny-tim said:
Hi sai2020! :smile:

Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) :smile:

I did that and I ended up with [tex]\int^1_{-1} log_e(\frac{1+u}{1-u})[/tex].

Is that correct?
 
  • #4
sai2020 said:
I did that and I ended up with [tex]\int^1_{-1} log_e(\frac{1+u}{1-u})[/tex].

Hi sai2020! :smile:

[tex]\int^1_{-1}[/tex] is correct …

but log((1+u)/(1-u)) = log((coshx + sinhx)/(coshx - sinhx)) = 2x,

and you haven't converted dx into du. :frown:

… but didn't the question only ask for the limits? :confused:
 
  • #5
Hi Tim! :)

Well this is a part of a bigger problem and I need to find g(u) as well. Here's what I did

Sorry I made a mistake. Is it

[tex] exp(\frac{-1}{2} (log_e \frac{1+u}{1-u})^2) [/tex]

How do I simplify further?

Thanks a lot :)
 
  • #6
tiny-tim said:
and you haven't converted dx into du. :frown:

yeah that would be [tex]dx = (1-u^2) du[/tex]
 
  • #7
sai2020 said:
yeah that would be [tex]dx = (1-u^2) du[/tex]

No … [tex]dx = \frac{du}{1-u^2}[/tex] :smile:
sai2020 said:
[tex] exp(\frac{-1}{2} (log_e \frac{1+u}{1-u})^2) [/tex]

How do I simplify further?

i suppose … [tex]\left(\frac{1\,+\,u}{1\,-\,u}\right)^{\frac{1}{2}\,log\frac{1\,+\,u}{1\,-\,u}}[/tex]

but I don't see where you go from there … :confused:

This isn't the usual way of solving [tex]\int^\infty_{-\infty} e^{{-x}^2} dx[/tex] :frown:
 
  • #8
Also wondering why on Earth do we want to use the substitution u=tanh(x) ?

If I could recall correctly, we evaluate the expression using either
i) gamma function
ii) normal distribution pdf
iii) polar coordinates.
 
  • #9
matematikawan said:
Also wondering why on Earth do we want to use the substitution u=tanh(x) ?

Hi matematikawan! :smile:

I think it's just an exercise in converting limits :smile:

which is, after all, what most people on this forum seem to find the difficult part of calculating an integral by substitution! :wink:
 

1. What is the purpose of using substitution in this integration?

Substitution is used to simplify the integrand and make it easier to integrate. In this particular case, the substitution u=tanh(x) helps to transform the integrand from a Gaussian function to a polynomial function, making it easier to evaluate.

2. How do you choose the appropriate substitution?

The choice of substitution is based on the structure of the integrand. In this case, the integral involves a Gaussian function and the substitution u=tanh(x) was chosen because it transforms the integrand into a polynomial function, which is easier to integrate.

3. What are the steps for integrating using substitution?

The steps for integrating using substitution are as follows:

  1. Choose an appropriate substitution that simplifies the integrand.
  2. Calculate the derivative of the substitution variable and substitute it in the integral.
  3. Simplify the integrand using the substitution.
  4. Integrate the simplified integrand.
  5. Substitute back the original variable in the final answer.

4. Can the substitution u=tanh(x) be used for all integrals involving exponential functions?

No, the substitution u=tanh(x) is specific to integrals involving Gaussian functions. Different substitutions may be required for other types of exponential functions.

5. Are there any limitations to using substitution for integration?

Yes, there are certain types of integrals for which substitution may not work. In such cases, other integration techniques, such as integration by parts or partial fractions, may be used.

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