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

[sai2020]

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?

 

[tiny-tim]

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!

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

 

[sai2020]

Hi sai2020!

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

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

Is that correct?

 

[tiny-tim]

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

Hi sai2020!

\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.

… but didn't the question only ask for the limits?

 

[sai2020]

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 :)

 

[sai2020]

and you haven't converted dx into du.

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

 

[tiny-tim]

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

No … dx = \frac{du}{1-u^2}

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 …

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

 

[matematikawan]

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.

 

[tiny-tim]

Also wondering why on Earth do we want to use the substitution u=tanh(x) ?

Hi matematikawan!

I think it's just an exercise in converting limits …

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