Determine $\int \limits_0^{\infty} \frac1{x^4+1}dx$ 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...
 A: Hint: Also include the line from $0$ to $Ri$.
A: 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!
