# Contour integration + paramerization of $\frac{1}{x^4+1}$

there is this question use contour integration to calculate . Let and consider the following loop contour In the limit RA →∞, derive expressions for $\int_Af(z)dz$$,$$\int_Bf(z)dz$$,\int_Cf(z)dz in terms of the real integral I defined above. Hence, use the residue theorem to find I. My attempts: I managed to use residue theorem to get (ipisqrt(2))/4 - (pisqrt(2))/4 as e^(ipi/4) is the only pole that lies in this loop. Now I know (ipisqrt(2))/4 - (pisqrt(2))/4 = \int_Af(z)dz$$+$$\int_Bf(z)dz$$+\int_Cf(z)dz$ where
$\int_Cf(z)dz$ = .

I tried to parameterize the 1st and 2nd term of the right hand side using z=re^(itheta) but it ends up in a very complicated expression which I have no idea how to proceed from there. So what should I do to find the expressions for the 1st and 2nd terms of the right hand side?? Thanks

• Btw the final answer should be pisqrt(2)/4 Commented Oct 27, 2017 at 15:38

For the contour labelled $A$, we use the parameterization $z=Re^{i\phi}$. If $R>1$, then we have

\begin{align} \left|\int_A \frac1{1+z^4}\,dz\right|&=\left|\int_0^{\pi/2} \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi\right|\\\\ &\le \int_0^{\pi/2}\left|\frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\right|\,d\phi\\\\ &=\int_0^{\pi/2}\frac{R}{|R^4-1|}\,d\phi\\\\ &= \int_0^{\pi/2}\frac{R}{R^4-1}\,d\phi\\\\ &=\frac{\pi R/2}{R^4-1}\to 0 \,\,\text{as}\,\,R\to \infty \end{align}

where we used the triangle inequality $|z^4+1|\ge ||z^4|-1|$

For the contour $B$, note that $z=iy$ so that

\begin{align} \int_B \frac{1}{z^4+1}\,dz&=\int_R^0 \frac{1}{1+y^4}\,i\,dy\\\\ &=-i \int_0^R \frac1{1+x^4}\,dx\to -i\int_0^\infty \frac{1}{x^4+1}\,dx\,\,\text{as}\,\,R\to \infty \end{align}

Can you finish now?

• hi @Mark Viola , thanks for helping, really appreciate your help Commented Oct 27, 2017 at 16:12
• @john You're welcome. My pleasure. Commented Oct 27, 2017 at 16:30
• Hi @Mark Viola, can i still ask for the contour labelled A, why is R>1 use instead of R>0? I mean is it wrong if i use R>0? Commented Oct 28, 2017 at 2:09
• Also for segment A, is it necessary to include the triangle equality part i mean afterall R/|R^4+1| still goes to zero as R goes to inf? Commented Oct 28, 2017 at 2:12
• We restrict $R>1$ so that $|R^4-1|=R^4-1$. Commented Oct 28, 2017 at 2:17

an other way is $$x^4+1+2x^2-2x^2=(x^2+1)^2-2x^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$$

• hi @Dr. Sonnhard Graubner , thanks for your reply but I need to parametrize to get an expression for segment A and B which I run out of ideas... Commented Oct 27, 2017 at 15:43
• This doesn't address the OP's question - at all. Commented Oct 27, 2017 at 15:57

Try instead using the entire upper half plane rather than just the first quadrant, then exploit symmetry of the integrand on the real line. This avoids having to integrate along path $B$. You need to calculate an additional residue, but this is easy. Then you justify that the integral along path $C$ vanishes as the radius tends to infinity, leaving you with the desired integrand.

• That way forward works, of course, but is far less efficient. Note that the integral along $B$ is just $-i$ times the integral along $A$. And $1-i=\sqrt 2 e^{-i\pi/4}$. ;-)) Commented Oct 27, 2017 at 16:07
• hi @heropup, thanks for helping but unfortunately this question is restricted to only first quadrant otherwise I would have use your approach to this question, still really grateful for your help! Commented Oct 27, 2017 at 16:13