complex analysis - $int_0^infty frac{cos(tx)}{(x^2 - 2x + 2)},mathrm{d}x$ for $t$ real

This was a question on an old prelim exam in complex analysis: compute

$$\int_0^\infty \frac{\cos(tx)}{x^2 - 2x + 2}\,\mathrm{d}x$$

for $t$ real. I've tried…

• Residue calculus—it's easy to integrate the similar $\int_0^\infty \frac{\cos(tx)}{x^2+2} \mathrm{d}x$ largely because the integrand is even, but this integrand isn't. Similarly $\int_0^\infty \frac{\sin(tx)}{x^2+2} \mathrm{d}x$ seems hard. Even if the original integral was from $1$ to $\infty$, so the denominator was even about $x=1$, we seem to need this latter, hard integral involving $\sin(tx)$.


• Mathematica—even for $t=1$, it gives the answer in terms of $$\int_0^z \frac{\sin(t)}{t}\,\mathrm dt \quad \text{and}\quad \int_0^z \dfrac{\cos(t)}{t}\,\mathrm dt,$$ which is not helpful.


• Looked through Gamelin's Complex Analysis text for inspiration; everything close used even or odd integrands.



• Googling/searching here, though it's hard to search for such a specific type of integral.

There's a chance there's just a typo on the old prelim, for what it's worth.

Answer

The integral is even in $t$, so I will assume $t\ge0$.
$$
\begin{align}
&\int_0^\infty\frac{\cos(tx)}{x^2-2x+2}\,\mathrm{d}x\\
&=\int_{-1}^\infty\frac{\cos(tx+t)}{x^2+1}\,\mathrm{d}x\\

&=\frac1{2i}\int_{-1}^\infty\left(\frac1{x-i}-\frac1{x+i}\right)\cos(tx+t)\,\mathrm{d}x\\
&=\frac1{2i}\int_{-1-i}^\infty\frac{\cos(tx+t+it)}{x}\,\mathrm{d}x\\
&-\frac1{2i}\int_{-1+i}^\infty\frac{\cos(tx+t-it)}{x}\,\mathrm{d}x\\
&=\frac1{2i}\cos(t+it)\int_{-1-i}^\infty\frac{\cos(tx)}{x}\,\mathrm{d}x
-\frac1{2i}\sin(t+it)\int_{-1-i}^\infty\frac{\sin(tx)}{x}\,\mathrm{d}x\\
&-\frac1{2i}\cos(t-it)\int_{-1+i}^\infty\frac{\cos(tx)}{x}\,\mathrm{d}x
+\frac1{2i}\sin(t-it)\int_{-1+i}^\infty\frac{\sin(tx)}{x}\,\mathrm{d}x\\
&=-\frac1{2i}\cos(t+it)\mathrm{Ci}(-t-it)
-\frac1{2i}\sin(t+it)\left(\frac\pi2-\mathrm{Si}(-t-it)\right)\\
&\hphantom{=}+\frac1{2i}\cos(t-it)\mathrm{Ci}(-t+it)

+\frac1{2i}\sin(t-it)\left(\frac\pi2-\mathrm{Si}(-t+it)\right)
\end{align}
$$
This matches what Mathematica computes.

If the integral was over the entire real line, the answer would avoid $\mathrm{Ci}$ and $\mathrm{Si}$. In fact, using contour integration, we get
$$
\begin{align}
\int_{-\infty}^\infty\frac{\cos(tx)}{x^2-2x+2}\,\mathrm{d}x
&=\int_{-\infty}^\infty\frac{\cos(tx+t)}{x^2+1}\,\mathrm{d}x\\

&=\cos(t)\int_{-\infty}^\infty\frac{\cos(tx)}{x^2+1}\,\mathrm{d}x
\color{#C00000}{-\sin(t)\int_{-\infty}^\infty\frac{\sin(tx)}{x^2+1}\,\mathrm{d}x}\\
&=\cos(t)\,\mathrm{Re}\left(\int_{-\infty}^\infty\frac{e^{itx}}{x^2+1}\,\mathrm{d}x\right)\color{#C00000}{-0}\\
&=\cos(t)\,\mathrm{Re}\left(\frac1{2i}\int_\gamma\left(\frac1{x-i}-\frac1{x+i}\right)e^{itx}\,\mathrm{d}x\right)\\
&=\cos(t)\,\mathrm{Re}\left(\frac1{2i}\int_\gamma\frac{e^{itx}}{x-i}\,\mathrm{d}x-\frac1{2i}\int_\gamma\frac{e^{itx}}{x+i}\,\mathrm{d}x\right)\\
&=\cos(t)\,\mathrm{Re}\left(\frac1{2i}2\pi i e^{-t}-0\right)\\[6pt]
&=\pi\cos(t)\,e^{-t}
\end{align}
$$
Where $\gamma$ is the contour along the real axis and circling back counter-clockwise around the upper half-plane.

This may have been the intended question.