How Do You Solve This Integral Involving W(x) and W1(x)?

[Char. Limit]

This isn't a homework problem. I challenged myself to see if I could make this work... but I couldn't. Now I'm asking for help on it.

The question is simple.

$$\int \frac{W(x) W_1(x)}{x} \left(\frac{1}{W(x)+1} + \frac{1}{W_1(x)+1}\right) dx$$

Now I have what this equals right here (W(x) W_1(x) + C), but I don't know how to get there. Can anyone help me out?

 

[╔(σ_σ)╝]

Are there any restrictions? Certainly, this expression is not always integrable.

 

[Char. Limit]

Are there any restrictions? Certainly, this expression is not always integrable.

Well, from what I can tell, the function isn't defined at x=0 or x=-e^(-1). So, I suppose we can restrict the domain to (0, infinity)?

 

[╔(σ_σ)╝]

Well, from what I can tell, the function isn't defined at x=0 or x=-e^(-1). So, I suppose we can restrict the domain to (0, infinity)?

Where did you get the -e^(-1) from?
My concern is not just the singularities of x but the actual functions w(x) and w1(x).
There might not be defined or even riemann integrable.

 

[Char. Limit]

Where did you get the -e^(-1) from?
My concern is not just the singularities of x but the actual functions w(x) and w1(x).
There might not be defined or even riemann integrable.

Well, here's the article on the W function, if it helps...

I got the -e^(-1) because 1/(W(x)+1) is singular at the point x=-1/e. I was certain that the functions are defined, and believe that they are integrable, because I found the integrand by taking a derivative.

 

[╔(σ_σ)╝]

Well, here's the article on the W function, if it helps...

I got the -e^(-1) because 1/(W(x)+1) is singular at the point x=-1/e. I was certain that the functions are defined, and believe that they are integrable, because I found the integrand by taking a derivative.

This makes a lot more sense now!
I was not aware of this function or notation so the integral seemed strange to me. Thanks for claring that up.

I will look into this tomorrow after my finals. :-)

 

[awkward]

Try verifying that the derivative of

$$W(x) \; W_1(x) + C$$

is the integrand.

 

[Mute]

This isn't a homework problem. I challenged myself to see if I could make this work... but I couldn't. Now I'm asking for help on it.

The question is simple.

$$\int \frac{W(x) W_1(x)}{x} \left(\frac{1}{W(x)+1} + \frac{1}{W_1(x)+1}\right) dx$$

Now I have what this equals right here (W(x) W_1(x) + C), but I don't know how to get there. Can anyone help me out?

Well, the derivative of W(x) is

$$\frac{dW(x)}{dx} = \frac{W(x)}{x(1+W(x))}$$

so what your integral is is:

$$\int dx~\left( \frac{W(x)}{x(1+W(x))}W_1(x) + W(x) \frac{W_1(x)}{x(1+W_1(x))}\right)$$
$$=\int dx~\left( \frac{dW(x)}{dx}W_1(x) + W(x) \frac{dW_1(x)}{dx}\right) = \int dx \frac{d(W(x)W_1(x))}{dx} = W(x)W_1(x) + C$$

Since you got the integrand by taking a derivative, this is really just reversing the steps, but I don't know what else you would be looking for in terms of solving this integral.