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

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Discussion Overview

The discussion revolves around the integration of the function e^(-x^2) using the substitution u = tanh(x). Participants explore the conversion of infinite limits to finite limits and the subsequent simplification of the integral.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to convert the limits of the integral \int^\infty_{-\infty} e^{-x^2} dx to finite limits using the substitution u = tanh(x).
  • Another participant provides a hint regarding the definition of tanh(x) in terms of exponential functions.
  • A participant claims to have derived the integral \int^1_{-1} log_e((1+u)/(1-u)), questioning its correctness.
  • One participant confirms the limits are correct but points out that the conversion of dx into du has not been addressed.
  • Another participant expresses the need to find g(u) as part of a larger problem and proposes a new expression involving the logarithm.
  • There is a disagreement regarding the correct expression for dx in terms of du, with conflicting claims about the relationship between the two differentials.
  • One participant questions the rationale behind using the substitution u = tanh(x) and suggests alternative methods for evaluating the integral.
  • Another participant speculates that the exercise is primarily focused on the challenge of converting limits in integration.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the derived expressions and the conversion of differentials. There is no consensus on the best approach to solving the integral or the necessity of the substitution used.

Contextual Notes

Some participants note that the discussion may be part of a larger problem, indicating that assumptions about the context and scope of the integral may not be fully addressed.

sai2020
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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).

How do I go about it?
 
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sai2020 said:
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).

How do I go about it?

Hi sai2020! :smile:

Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) :smile:
 
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 \int^1_{-1} log_e(\frac{1+u}{1-u}).

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

Hi sai2020! :smile:

\int^1_{-1} 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:
 
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

exp(\frac{-1}{2} (log_e \frac{1+u}{1-u})^2)

How do I simplify further?

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

yeah that would be dx = (1-u^2) du
 
sai2020 said:
yeah that would be dx = (1-u^2) du

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

How do I simplify further?

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

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

This isn't the usual way of solving \int^\infty_{-\infty} e^{{-x}^2} dx :frown:
 
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.
 
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:
 

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