Difficult Improper Integral Involving Arctan(x)

[caleb5040]

So recently I've been working through some challenge problems from my old calculus textbook for fun. I'm stuck on one of the integrals, though, and can't find any solutions online. This isn't for homework...it's for my interest and hopefully the interest of others. Here it is (sorry about the formatting):

int((arctan(Pi*x)-arctan(x))/x, x = 0 .. infinity)

In the problem statement it says you first need to express the integral as an iterated integral. I also know the answer is (π/2)ln(π).

 

[SammyS]

So recently I've been working through some challenge problems from my old calculus textbook for fun. I'm stuck on one of the integrals, though, and can't find any solutions online. This isn't for homework...it's for my interest and hopefully the interest of others. Here it is (sorry about the formatting):

int((arctan(Pi*x)-arctan(x))/x, x = 0 .. infinity)

In the problem statement it says you first need to express the integral as an iterated integral. I also know the answer is (π/2)ln(π).

Interesting problem ...

\displaystyle \int_0^\infty \frac{\arctan(\pi x)-\arctan(x)}{x}\,dx

That this may come from an iterated integral suggests:

\displaystyle <br /> \int_0^\infty \left(\displaystyle \left.\arctan(y)\right|_{y=x}^{y=\pi x}\right)\frac{1}{x}\,dx​

arctan(y) is the anti-derivative of what function?

If you write this as an iterated integral, what happens if you change the order of integration ?

 

[Dickfore]

The answer is zero.

Edit:
Actually, because each term:
<br /> \int_{0}^{\infty}{\frac{\arctan x}{x} \, dx}<br />
is logarithmically divergent on the upper bound, we need to be more careful. Your integral reduces to the following limit:
<br /> \lim_{B \rightarrow \infty} \int_{B}^{B \pi}{\frac{\arctan x}{x} \, dx}<br />
[strikeout]Make the substitution x = \tan(z/2). You will get:
<br /> 2 \,\int^{2\arctan(B/\pi)}_{2\arctan{B}}{\frac{z}{\sin z} \, dz}, \ B \rightarrow \infty<br />
[/strikeout]

Edit2:
Scratch the last subst. Try this x = B y. You get:
<br /> \int_{1}^{\pi}{\frac{\arctan(B \, y)}{y} \, dy}, \ B \rightarrow \infty<br />
Now, in the above limit the arctangent is \pi/2, and the remaining integral is \ln \pi. Combining everything, we get:
<br /> \frac{\pi \, \ln \pi}{2}<br />

 

[caleb5040]

Awesome! Thank you both. Those were just the thoughts I needed to do it myself.
Sorry for not posting my initial attempts...they both involved rewriting the integral as an iterated integral and switching the order (I figured out two ways of doing this). The trouble was that each time the integral became more complex. Now it makes sense, though.

 

[Boorglar]

You might be interested to know that in general, if \frac{f(x)}{x} is integrable on any interval [a , b] for 0&lt;a&lt;b then for all \alpha, \beta &gt; 0:

\int_{0}^{∞} \frac{f(\alpha x)-f(\beta x)} {x} dx = (A-B)\ln(\frac{\beta} {\alpha})

where \displaystyle A = \lim_{x \rightarrow 0^{+}} f(x) and \displaystyle B = \lim_{x \rightarrow ∞} f(x)

 

[caleb5040]

You might be interested to know that in general, if \frac{f(x)}{x} is integrable on any interval [a , b] for 0&lt;a&lt;b then for all \alpha, \beta &gt; 0:

\int_{0}^{∞} \frac{f(\alpha x)-f(\beta x)} {x} dx = (A-B)\ln(\frac{\beta} {\alpha})

where \displaystyle A = \lim_{x \rightarrow 0^{+}} f(x) and \displaystyle B = \lim_{x \rightarrow ∞} f(x)

Interesting! I think I see how to prove the generalization.