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

1. Sep 6, 2008

### 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?

Last edited: Sep 6, 2008
2. Sep 6, 2008

### tiny-tim

Hi sai2020!

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

3. Sep 6, 2008

### sai2020

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

Is that correct?

4. Sep 6, 2008

### tiny-tim

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?

5. Sep 6, 2008

### 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 :)

6. Sep 6, 2008

### sai2020

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

7. Sep 6, 2008

### tiny-tim

No … $$dx = \frac{du}{1-u^2}$$
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 $$\int^\infty_{-\infty} e^{{-x}^2} dx$$

8. Sep 7, 2008

### 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.

9. Sep 8, 2008

### tiny-tim

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!