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

  • Thread starter sai2020
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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
tiny-tim
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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?
Hi sai2020! :smile:

Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) :smile:
 
  • #3
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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
tiny-tim
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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
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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
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and you haven't converted dx into du. :frown:
yeah that would be [tex]dx = (1-u^2) du[/tex]
 
  • #7
tiny-tim
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yeah that would be [tex]dx = (1-u^2) du[/tex]
No … [tex]dx = \frac{du}{1-u^2}[/tex] :smile:
[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
tiny-tim
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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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