Determine $\int \limits_0^{\infty} \frac1{x^4+1}dx$

Question

Let $$f(z)=\frac1{1+z^4}$$ (a) Find the sinularity of $f(z)$ in the first quadrant where $Re(z), Im(z) \ge 0$.

(b) Find the residue of the singular point found in the first quadrant.

(c) Let $\Gamma_R$ be the quarter circle $\Gamma_R: |z|=R$, $Re(z), Im(z) \ge 0$, positively oriented. Show that $$\lim \limits_{R \rightarrow \infty} \int _{\Gamma_R} f(z) \, \, dz =0$$

(d) Determine $$\int \limits_0^{\infty} \frac1{x^4+1}dx$$

Attempt:

For (a) I got $$z_0 = \frac{\sqrt2}{2} +\frac{\sqrt2}{2}i$$

For (b) I got that this is a simple pole since the degree of is $1$ and $f(z_0)$ doesn't make the numerator vanish. So using the limit formula $$Res (f,z_0)= -\frac{\sqrt2}{8}-\frac{\sqrt2}{8}i$$

Used $ML$ Lemma for (c).

Stuck on (d). What should i make my region? I was thinking $\Gamma = \Gamma_R + \nabla R$ where $\Gamma_R$ is as stated in the question and $\nabla_R$ is the line from $0$ to $R$ in the real axis. But this would mean $\Gamma $ is not closed. Is this a problem?

From using this i got the answer to (d) as $$\frac{\pi \sqrt2}4 -\frac{\pi \sqrt2}4i$$ but really unsure on this since my chosen region is not closed...

u should use a quartercircle containing the positive half line, the part of the imaginary axis with $Im[z]\geq 0$ and the arc connecting them. — tired, May 27 '15 at 13:58

score 1 · Answer 1 · answered May 27 '15 at 13:56

1

Hint: Also include the line from $0$ to $Ri$.

answered May 27 '15 at 13:56

Tim Raczkowski

20,046

That would be oriented downwards right? to be consistent with the anticlockwise direction – cooldudeman May 27 '15 at 14:12
As in from $Ri$ to $0$? – cooldudeman May 27 '15 at 14:13
Yes, and then let $R\to\infty$. – Tim Raczkowski May 27 '15 at 14:13
OK let $\Gamma = \Gamma_R + \nabla_R + \rho _R$ where the last one is the one we just stated. But what would $$\int _{\rho_R } f(z) dz$$ even be? As R tends to infinity? – cooldudeman May 27 '15 at 14:16
Well for $R$ large enough you'll get $2\pi i$ Res. See tired's answer below. – Tim Raczkowski May 27 '15 at 14:19
1

$\Gamma_R "=" [0,R]+[iR,0]+[Re^{i\phi},\phi \in [0, \pi/2]]$ – tired May 27 '15 at 14:21

tired · Accepted Answer · 2015-05-27T15:06:40.923

There are two straightforward possibilities:

1.) Take the Quartercircle in the first Quadrant

Every other quadrant should work as well, but maybe there are a few more nasty minus signs. You can now show that the integrals along the real and imaginary axis are equal (up to a prefactor).:

$ I_{im}=\int_{0}^{i \infty}\frac{1}{1+z^4}dz=i \int_{0}^{ \infty}\frac{1}{1+(i t)^4}dt=i \int_{0}^{ \infty}\frac{1}{1+ t^4}dt=i I_{re} $

Now applying residue theorem, we obtain $$ I_{re}-iI_{re}=2\pi i \times \text{res}\left[z_0=\frac{1}{2\sqrt{2}}\left(1+i\right)\right]=\frac{\pi}{2\sqrt{2}}\left(1-i\right) $$

Note the minus sign which comes from the fact that we go from $i\infty$ to $0$.

Therefore: $$ I_{re}=\frac{\pi}{2\sqrt{2}} $$

2.) Take a half circle in the upper/(lower) half plane and divide by two

Applying symmetry w.r.t to $z \rightarrow -z$,

$$2 I_{re} = 2\pi i \times \text{res}\left[z_0=\frac{1}{2\sqrt{2}}\left(1+i\right)\right]+2\pi i \times \text{res}\left[z_0=\frac{1}{2\sqrt{2}}\left(1-i\right)\right]$$

Note that we now need both residues in the upper half plane. The rest of the calculation is now some easy algebra and yields the same answer as above!

So I can make my region the upper semi circle with radius R centred at 0 and the line from -R to R on the real line to find $$\int \limits_{-\infty}^{\infty} \frac1{x^4+1}dx$$ and then divide by $2$ to get the final answer? — cooldudeman, May 27 '15 at 14:43

Determine $\int \limits_0^{\infty} \frac1{x^4+1}dx$

2 Answers2