The Proof is Trivial!

Problem 546 (*/**)

Evaluate $\displaystyle I_n = \int_0^\frac{\pi}{2} \frac{1}{1+\tan^n(x)} \ dx$

Solution 546:

$I_n=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{1+\tan^n(x)} \ dx = \displaystyle\int_0^{\frac{\pi}{2}} \dfrac{\cos^n(x)}{\sin^n(x)+\cos^n(x)} \ dx$
Sending $x \mapsto \dfrac{\pi}{2}-x$ yields:
$I_n=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{\sin^n(x)}{\sin^n(x)+\cos^n(x)} \ dx[br]\Rightarrow 2I_n=\displaystyle\int_0^{\frac{\pi}{2}} 1 \ dx[br]\Rightarrow I_n=\dfrac{\pi}{4}$.

Thanks, someone did it in latex. I really need to learn how to use this..


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It ain't all that hard although on TSR it can be quite buggy at times.

Unless ive missed something terribly obvious 😁 i havent done integration in a while.. Since step III actually.


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Yes, that's fine. It's very straightforward if you know what to do, but it's a pretty nice result anyway.

Nicely done, and I don't blame you for that second residue

Besides finding the second derivative, another way of calculating it would involve considering a small circle around the origin. The residue would then be the coefficient of $z^2$ in the expansion of

$\dfrac{1}{(2z-1)(z-2)}$

but that's hardly mess-free

Very pretty.

As I guess you know, that substitution is standard for integrals involving sin and cos between 0 and 2pi.

And as for the second residue, yes, I simply brute forced it rather than thinking about any cleaner approach. And speaking of cleaner approaches ..

.. I'm not sure I'm following you. Where does the coefficient of z^2 come into the picture? I'm trying to imagine a Laurent series method here, but that doesn't seem to be what you're doing.

Well here's my method:

Problem 548**

By considering a suitable integral, or otherwise, prove that

$\ln 2 < \frac{\pi^2}{12} < 1$.

A proof using series, because integrals are boring. .

Problem 550

Show that

$\displaystyle \int_{0}^{\infty} \dfrac{\ln^3 x}{x^2 + 1} \ dx = 0$

Solution:



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