MHB Evaluate the integral ∫[arctan(ax)−arctan(bx)]/xdx

[lfdahl]

Evaluate the integral:

\[ \int_{0}^{\infty}\frac{\arctan(ax)-\arctan(bx)}{x}dx\]

where $a,b \in \mathbb{R}_+$

 

[Euge]

It's been a while since I've done one of these challenges. :D Here is my solution.

Spoiler

The answer is $\displaystyle \frac{\pi}{2}\ln\frac{a}{b}$. Note

$$\arctan(ax) - \arctan(bx) = \int_b^a \frac{x}{1 + (tx)^2}\, dt$$

Fubini's theorem allows us to write

$$\int_0^\infty \frac{\arctan(ax) - \arctan(bx)}{x}\, dx = \int_b^a \int_0^\infty \frac{dt}{1 + (tx)^2}\, dx$$

This is justified as

$$\int_1^\infty \frac{dt}{1 + (tx)^2} < \int_1^\infty \frac{dt}{t^2x^2} = \frac{1}{x^2}$$ and $\displaystyle \int_b^a \frac{dx}{x^2} < \infty$.

Now

$$\int_0^\infty \frac{dt}{1 + (tx)^2} = \frac{\arctan(tx)}{x}\bigg|_{t = 0}^\infty = \frac{\pi}{2x}$$ and $\displaystyle\int_b^a \frac{\pi}{2x}\, dx = \frac{\pi}{2}\ln\frac{a}{b}$. Therefore

$$\int_0^\infty \frac{\arctan(ax) - \arctan(bx)}{x}\, dx = \frac{\pi}{2}\ln\frac{a}{b}$$

 

[lfdahl]

It's been a while since I've done one of these challenges. :D Here is my solution.

Spoiler

The answer is $\displaystyle \frac{\pi}{2}\ln\frac{a}{b}$. Note

$$\arctan(ax) - \arctan(bx) = \int_b^a \frac{x}{1 + (tx)^2}\, dt$$

Fubini's theorem allows us to write

$$\int_0^\infty \frac{\arctan(ax) - \arctan(bx)}{x}\, dx = \int_b^a \int_0^\infty \frac{dt}{1 + (tx)^2}\, dx$$

This is justified as

$$\int_1^\infty \frac{dt}{1 + (tx)^2} < \int_1^\infty \frac{dt}{t^2x^2} = \frac{1}{x^2}$$ and $\displaystyle \int_b^a \frac{dx}{x^2} < \infty$.

Now

$$\int_0^\infty \frac{dt}{1 + (tx)^2} = \frac{\arctan(tx)}{x}\bigg|_{t = 0}^\infty = \frac{\pi}{2x}$$ and $\displaystyle\int_b^a \frac{\pi}{2x}\, dx = \frac{\pi}{2}\ln\frac{a}{b}$. Therefore

$$\int_0^\infty \frac{\arctan(ax) - \arctan(bx)}{x}\, dx = \frac{\pi}{2}\ln\frac{a}{b}$$

Excellent, Euge! Thankyou very much for your participation!