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

• tolove
In summary: However, the notation used is somewhat unconventional, and could be interpreted as the second one. It would be best to clarify with the person who wrote the original conversation.
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?

tolove said:
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.

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

tolove said:
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
$$\int_0^{\infty} \frac{1}{(1 + x^2)(1 + \arctan(x)}\, dx ?$$

Ray Vickson said:
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.

## 1. What is an improper integral?

An improper integral is an integral that either has an infinite limit of integration or has a function that is not defined at one or more points within the limits of integration. This means that the area under the curve cannot be calculated using the standard methods and requires a different approach.

## 2. How is an improper integral evaluated?

An improper integral is evaluated by taking the limit of a definite integral as one or both of the limits of integration approach infinity or a point of discontinuity. The resulting value is the approximation of the area under the curve.

## 3. What are the types of improper integrals?

There are two types of improper integrals: Type 1 and Type 2. Type 1 improper integrals have an infinite limit of integration or a point of discontinuity within the limits, while Type 2 improper integrals have both limits of integration equal to infinity.

## 4. How do I know if an integral is improper?

An integral is improper if it meets at least one of the following conditions: the limit of integration is infinite or the function is not defined at one or more points within the limits. If either of these conditions is met, the integral cannot be evaluated using the standard methods and is considered improper.

## 5. What are some common techniques for evaluating improper integrals?

Some common techniques for evaluating improper integrals include using the limit definition, breaking the integral into smaller pieces, and using substitution or integration by parts. Additionally, some improper integrals can be evaluated using the comparison test or the L'Hopital's rule.

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