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Given an integral, $\int_{-\infty}^\infty f(t) \,dt$, where $f(t)=\frac{1}{t^4+1}$, we wish to evaluate the integral without resorting the the residue theorem, but using Cauchy's Integral Formula only. The C.I.F. can be stated as:

Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0)=\frac{1}{2\pi i}\int_C \frac{f(z)\,dz}{z-z_0}. $$

I need a hint or somewhere to get started. I can let $\int_{-\infty}^\infty\frac{1}{z^4+1} \, dz$, for which I can write out either $\int_{-\infty}^\infty \frac{1}{z^2-1}\cdot\frac{1}{z^2+1} \, dz$ or perhaps $\int_{-\infty}^\infty \frac{z-\sqrt{2}}{2\sqrt{2}(-(z)^2 + \sqrt{2}z-1)} \cdot \frac{z+\sqrt{2}}{2\sqrt{2}(z^2+\sqrt{2}z+1)} \, dz$, via partial fraction decomposition. For the decomposition, I know that $f(z)$ has zeros at $\sqrt{2}$ and $-\sqrt{2}$, respectively. This really impedes what I want to do, since there will be two singular points on the contour I will want to draw. I'm not sure how to get around this. I suppose for each point I could use the "keyhole" method, but I'm unsure.

I'm also unsure how I'm supposed to use the C.I.F. once I were to get a closed contour. Lets assume the keyhole method works for the singular points, then I have a contour $C_1$ from $-R$ to $R$ and another contour $C_2$ connecting via $Re^{it}$, with $0\leq t \leq \pi$. Rewriting the C.I.F. I will need to show some things and have $f(z_0)\cdot 2\pi i(z-z_0) = \int_Cf(z) \, dz = \int_{C_1}f(z) \, dz+\int_{C_2}f(z) \, dz$. I'm really not sure whats going on here... I do know that the integral of $C_2$ will be zero has $R$ goes to infinity, assuming keyhole method works, by the Cauchy-Groursat theorem, but thats a big "if".

Also, why are my pi's missing from the LaTeX?

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  • $\begingroup$ If you want to evaluate $\int_{\Bbb R}\frac{dt}{t^4+1}=2\int_0^\infty\frac{dt}{t^4+1}$, you can even do without the CIF if you use $t=\tan^{1/2}\theta$ and basic identities involving Beta and Gamma functions. (In fact, with $t=\tan^{2/n}\theta$ you can prove $\int_0^\infty\frac{dt}{t^n+1}=\frac{1}{\operatorname{sinc}\frac{\pi}{n}}$ for $n>1$.) $\endgroup$ – J.G. Oct 14 '19 at 13:04
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    $\begingroup$ Sounds familiar... $\endgroup$ – David C. Ullrich Oct 14 '19 at 13:52
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This is not an answer. It happens that I'm the guy who assigned the question; the point to this non-answer is to clarify the point.

The point was to evaluate the integral specifically using the Cauchy Integral Formula, by analogy with the following evaluation of $\int1/(t^2+1)$:

Let $f(z)=\frac1{z^2+1}$. Let $C_R$ be the usual semicircle. If we define $g(z)=\frac1{z+i}$ then CIF shows that $$\int_{C_R}f=\int_{C_R}\frac{g(z)}{z-i}\,dz=2\pi i g(i)=\pi\quad(R>1).$$Now let $R\to\infty$, etc.

One little added trick works for $1/(t^4+1)$. No explicit partial fractions, no keyhole contours...

Hint: $C_R=C_R^1\dot+ C_R^2$. Also $z^4+1=\prod_{k=1}^4(z-\omega_k)$.

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