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

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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tiny-tim
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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?
Hi sai2020! Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + e-x) Hi sai2020! Hint: tanhx = sinhx/coshx = (ex - e-x)/(ex + 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
Homework Helper
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? 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 :)

and you haven't converted dx into du. yeah that would be $$dx = (1-u^2) du$$

tiny-tim
Homework Helper
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 $$\int^\infty_{-\infty} e^{{-x}^2} dx$$ 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
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! 