- #1

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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).

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- Thread starter caleb5040
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- #1

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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).

- #2

SammyS

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Interesting problem ...

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).

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

That this may come from an iterated integral suggests:

[itex]\displaystyle

\int_0^\infty \left(\displaystyle \left.\arctan(y)\right|_{y=x}^{y=\pi x}\right)\frac{1}{x}\,dx[/itex]

\int_0^\infty \left(\displaystyle \left.\arctan(y)\right|_{y=x}^{y=\pi x}\right)\frac{1}{x}\,dx[/itex]

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 ?

- #3

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The answer is zero.

Edit:

Actually, because each term:

[tex]

\int_{0}^{\infty}{\frac{\arctan x}{x} \, dx}

[/tex]

is logarithmically divergent on the upper bound, we need to be more careful. Your integral reduces to the following limit:

[tex]

\lim_{B \rightarrow \infty} \int_{B}^{B \pi}{\frac{\arctan x}{x} \, dx}

[/tex]

[strikeout]Make the substitution [itex]x = \tan(z/2)[/itex]. You will get:

[tex]

2 \,\int^{2\arctan(B/\pi)}_{2\arctan{B}}{\frac{z}{\sin z} \, dz}, \ B \rightarrow \infty

[/tex]

[/strikeout]

Edit2:

Scratch the last subst. Try this [itex]x = B y[/itex]. You get:

[tex]

\int_{1}^{\pi}{\frac{\arctan(B \, y)}{y} \, dy}, \ B \rightarrow \infty

[/tex]

Now, in the above limit the arctangent is [itex]\pi/2[/itex], and the remaining integral is [itex]\ln \pi[/itex]. Combining everything, we get:

[tex]

\frac{\pi \, \ln \pi}{2}

[/tex]

Edit:

Actually, because each term:

[tex]

\int_{0}^{\infty}{\frac{\arctan x}{x} \, dx}

[/tex]

is logarithmically divergent on the upper bound, we need to be more careful. Your integral reduces to the following limit:

[tex]

\lim_{B \rightarrow \infty} \int_{B}^{B \pi}{\frac{\arctan x}{x} \, dx}

[/tex]

[strikeout]Make the substitution [itex]x = \tan(z/2)[/itex]. You will get:

[tex]

2 \,\int^{2\arctan(B/\pi)}_{2\arctan{B}}{\frac{z}{\sin z} \, dz}, \ B \rightarrow \infty

[/tex]

[/strikeout]

Edit2:

Scratch the last subst. Try this [itex]x = B y[/itex]. You get:

[tex]

\int_{1}^{\pi}{\frac{\arctan(B \, y)}{y} \, dy}, \ B \rightarrow \infty

[/tex]

Now, in the above limit the arctangent is [itex]\pi/2[/itex], and the remaining integral is [itex]\ln \pi[/itex]. Combining everything, we get:

[tex]

\frac{\pi \, \ln \pi}{2}

[/tex]

Last edited:

- #4

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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.

- #5

- 210

- 10

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

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

- #6

- 22

- 0

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

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

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

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