(Improper Integrals) Not sure if I'm doing this properly

[tolove]

Initial improper integral:
∫ dx / (1+x**2) * (1+ atan(x)) , x = 0, ∞

Substitutions:
μ = 1 + atan(x)
dμ = dx / (1 + x**2)

μ(∞) = 1 + pi/2
μ(0) = 1

Integral:
∫ dμ / μ , μ = 1, 1+ pi/2

Then solve.

I'm getting the right answer, but I think I'm botching something due to a lack of understanding about how notation works with indefinite integrals. Is this right?

 

[Mark44]

Initial improper integral:
∫ dx / (1+x**2) * (1+ atan(x)) , x = 0, ∞

Substitutions:
μ = 1 + atan(x)
dμ = dx / (1 + x**2)

μ(∞) = 1 + pi/2
μ(0) = 1

Integral:
∫ dμ / μ , μ = 1, 1+ pi/2

Then solve.

I'm getting the right answer, but I think I'm botching something due to a lack of understanding about how notation works with indefinite integrals. Is this right?

This is an improper integral, so you're going to need to evaluate a limit.

$$ \int_0^{\infty} \frac{(1 + arctan(x))dx}{1 + x^2}$$
$$ = \lim_{b \to \infty} \int_0^b \frac{(1 + arctan(x))dx}{1 + x^2}$$

The integral itself is fairly easy - you have the right substitution.

 

[STEMucator]

It is indeed an improper integral, and the fact you can find a finite value for it implies that the integral converges.

 

[Ray Vickson]

Initial improper integral:
∫ dx / (1+x**2) * (1+ atan(x)) , x = 0, ∞

Substitutions:
μ = 1 + atan(x)
dμ = dx / (1 + x**2)

μ(∞) = 1 + pi/2
μ(0) = 1

Integral:
∫ dμ / μ , μ = 1, 1+ pi/2

Then solve.

I'm getting the right answer, but I think I'm botching something due to a lack of understanding about how notation works with indefinite integrals. Is this right?

What you wrote means
$$\int_0^{\infty} \frac{1 + \arctan(x)}{1+x^2} \, dx.$$ Is that what you meant, or did you mean
$$\int_0^{\infty} \frac{1}{(1 + x^2)(1 + \arctan(x)}\, dx ?$$

 

[Mark44]

What you wrote means
$$\int_0^{\infty} \frac{1 + \arctan(x)}{1+x^2} \, dx.$$ Is that what you meant, or did you mean
$$\int_0^{\infty} \frac{1}{(1 + x^2)(1 + \arctan(x)}\, dx ?$$

Based on context, I would say that the OP meant the first one, above.