0

$f:(0,\frac{\pi}{2})\to$$\mathbb{R}$, $f(a) = \int_0^1\frac{dx}{x^2+2x\tan a + 1}$, $\forall a\in(0,\frac{\pi}{2})$

What is $\lim_ {a\to \frac{\pi}{4}} f(a)$?

I am confused. Function $f$'s domain is given at the beginning but then this $\forall a\in(0,\frac{\pi}{2})$ is given . Isn't $a$ the variable with which $f$ is defined?

The only thing i can extract from here is $\tan a$ being positive always but aside from that i do not know what to do. I tried writing the denominator by "completing the square" and then i used u-substitution but in the end i get $\frac{1}{2\sqrt {\tan^2{a -1}}}$ multiplied by some logarithms but this is defined only for ${\tan^2a} \gt 1$.

How do i solve it?

Pete42
  • 197
  • The easiest thing we can do to continue what you did so far is consider $\tan a> 1, \tan a< 1$ and $\tan a=1$ cases separately. It will give you $f$ as a piecewise function. – Bumblebee Nov 23 '20 at 16:27
  • @VictorPalea this is true if f is continuous but how do i show this ? – Pete42 Nov 23 '20 at 16:34
  • @Bumblebee do you suggest another way or should i just continue with what i did and use cases ? – Pete42 Nov 23 '20 at 16:35
  • @Bumblebee actually i think what you said is easy enough because i can let $\tan^2a-1$ for $\tan a \gt 1$ and $-(\tan^2a-1)$ for $\tan a \lt 1$. Thanks ! – Pete42 Nov 23 '20 at 16:40
  • I am glad, if my comment helped you. Now you can answer to your own question :) – Bumblebee Nov 23 '20 at 16:44
  • @Bumblebee Why is it wrong to let $a\to\pi/4$ before integrating and answer $$\int_0^1 \frac{1}{x^2+2 x+1} , dx=\frac{1}{2}$$Which is the result of the limit, BTW – Raffaele Nov 23 '20 at 17:06
  • @Raffaele: I didn't say it is wrong. But you need to know the continuity of $f$ at $\pi/4$ prior to do so. On the other hand, interchanging limits and integrals doesn't work nicely. This is one of the motivations for Lebesgue integration. – Bumblebee Nov 23 '20 at 17:22
  • You can prove that $g(b) = \int_0^1 dx/(x^2 + 2 b x + 1)$ is continuous at $b = 1$ directly by writing $g(1 + \epsilon) - g(1)$ as $\epsilon$ times an integral and showing that the integral is $O(1)$. Then you just need to evaluate $f(\pi/4)$. – Maxim Nov 24 '20 at 14:26

2 Answers2

0

We want to find the indefinite integral.

First we complete the square on the denominator($x^2+2x\tan a + 1$).

We get $(x+\tan a)^2 - (\tan^2 a - 1)$

Now this can be easily solved with u substitution with $x + \tan a$ but $(\tan^2 a - 1)$ it going to be a problem under the square root if it is negative in the form $\frac{1}{x^2-c^2}$ where $c$ would be $\sqrt{(\tan^2 a - 1)}$ .

So we take 3 cases :

-$\tan a \gt 1$ we continue with the previous solution

-$\tan a \lt 1$ we write $(\tan^2 a - 1)$ as $-(-\tan^2 a + 1)$ and solve with elementary integrals

-$\tan a = 1$ we get $(\tan^2 a - 1) = 0$ and solve with elementary integrals as well

After this it is easy.

Pete42
  • 197
0

Consider first $$x^2+2k x+1=(x+k)^2+(1-k^2)$$ $$I=\int \frac {dx}{x^2+2k x+1} =\int \frac {dx}{(x+k)^2+(1-k^2)}$$ Let $x+k=t\sqrt{1-k^2} $ $$I=\frac 1 {\sqrt{1-k^2}} \int \frac {dt}{t^2+1}=\frac{\tan ^{-1}(t)}{\sqrt{1-k^2}}$$

Back to $x$ $$I=\frac{\tan ^{-1}\left(\frac{x+k}{\sqrt{1-k^2}}\right)}{\sqrt{1-k^2}}$$ Using the bounds $$J=\int_0^1 \frac {dx}{x^2+2k x+1}=\frac{\tan ^{-1}\left(\frac{k+1}{\sqrt{1-k^2}}\right)-\tan ^{-1}\left(\frac{k}{\sqrt{1-k^2}}\right)}{\sqrt{1-k^2}}$$ Recombine the arctangents $$J=\frac{1}{\sqrt{1-k^2}}\tan ^{-1}\left(\frac{\sqrt{1-k}}{\sqrt{1+k}}\right)$$

Now, using Taylor series around $k=1^\pm$, in both cases $$J=\frac{1}{2}-\frac{1}{6}(k-1)+\frac{1}{15} (k-1)^2+O\left((k-1)^{3}\right)$$

Claude Leibovici
  • 260,315