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$$\int_0^{\infty} \frac{x^2+1}{x^4+1}dx$$

What i've found are the singularities at: $e^{\pi/4+\pi/2n}$ for $n=0,1,2,3$. But i'm unsure how to calculate the integral since I don't want to include the singularity at $x_2=\frac{-1+i}{\sqrt{2}}$ since then I would be calculating the integral from negative to positive infinity.

My idea was to construct a contour $\gamma : x=ti$ where $R<t\leq0$ and $\lim_{R\rightarrow \infty}$. My reckoning was that since:

$$\int_{\gamma}\frac{x^2+1}{x^4+1}dx=i\int_R^0\frac{1-t^2}{t^4+1}dt$$

doesn't have any singularities along the positive real numberline, it is therefore bounded and thus:

$$i\int_R^0\frac{1-t^2}{t^4+1}dt\leq \Big| i\int_R^0\frac{1-t^2}{t^4+1}dt\Big|\leq\int_R^0\frac{|1-t^2|}{|t^4+1|}dt\leq\frac{|R^2|+1}{|R^4|-1}\rightarrow0 \text{ when } R\rightarrow \infty.$$

The answer would then be: $$\lim_{R\rightarrow \infty}\int_0^{R}\frac{x^2+1}{x^4+1}dx=2\pi iRes(f;e^{\pi/4})-\int_0^{\pi/2}f(x)dx-i\int_R^0f(x)dx=\frac{\pi}{2\sqrt{2}}.$$

But sadly I am mistaken, what is wrong?

Strange Brew
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  • Hint The integrand is even, so the integral is equal to $\frac{1}{2} \int_{-\infty}^{\infty} \sim dx$. – Travis Willse May 07 '16 at 00:27
  • Thanks for the tip. That still won't add upp for me. With both residues I get: $2\pi i\frac{1}{2\sqrt{2}i}=\frac{\pi}{\sqrt{2}}$ which is correct but then half of that is still the same as above. – Strange Brew May 07 '16 at 00:53
  • Hm, I seem to get the correct answer. Notice that the last inequality in the second-to-last display equation simply isn't right, for a few reasons, though this isn't the source of the wrong answer, as that integral turns out to have zero value anyway. Did you perhaps forget to multiply by $2$ when you computed $2 \pi i \operatorname{Res}(f; e^{\pi i / 4})$? (Note also that $i$'s are missing from various exponents in the question.) – Travis Willse May 07 '16 at 10:24

1 Answers1

2

Let $I$ be the integral defined by

$$I=\int_0^\infty\frac{x^2+1}{x^4+1}\,dx \tag 1$$

We proceed to evaluate the integral in $(1)$ by analyzing the contour integral $J$

$$J=\oint_C \frac{z^2+1}{z^4+1}\,dz$$

where $C$ is the quarter circle in the first quadrant, centered at the origin, with radius $R$. Then, we have

$$J=\int_0^R \frac{x^2+1}{x^4+1}\,dx+i\int_0^R \frac{y^2-1}{y^4+1}\,dy+\int_0^{\pi/2}\frac{R^2e^{i2\phi}+1}{R^4e^{i4\phi}+1}\,iRe^{i\phi}\,d\phi \tag 2$$

From the residue theorem, we find that

$$\begin{align} J&=2\pi i \text{Res}\left(\frac{z^2+1}{z^4+1}, z=e^{i\pi/4}\right)\\\\ &=2\pi i \frac{e^{i\pi/2}+1}{4e^{i3\pi/4}}\\\\ &=\frac{\sqrt{2}\pi}{2} \end{align}$$

As $R\to \infty$, the third integral on the right-hand side of $(2)$ vanishes and we find

$$\int_0^\infty \frac{x^2+1}{x^4+1}\,dx+i\int_0^\infty \frac{y^2-1}{y^4+1}\,dy=\frac{\sqrt{2}\pi}{2} \tag 3$$

Equating the real parts of $(3)$, we find

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^2+1}{x^4+1}\,dx=\frac{\sqrt{2}\pi}{2}}$$

As a bonus, we find that

$$\int_0^\infty \frac{x^2-1}{x^4+1}\,dx=0$$

which one can verify directly by enforcing the substitution $x\to 1/x$.

Mark Viola
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