This was a question on an old prelim exam in complex analysis: compute
∫∞0cos(tx)x2−2x+2dx
for t real. I've tried…
- Residue calculus—it's easy to integrate the similar ∫∞0cos(tx)x2+2dx largely because the integrand is even, but this integrand isn't. Similarly ∫∞0sin(tx)x2+2dx seems hard. Even if the original integral was from 1 to ∞, 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 ∫z0sin(t)tdtand∫z0cos(t)tdt, 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≥0.
∫∞0cos(tx)x2−2x+2dx=∫∞−1cos(tx+t)x2+1dx=12i∫∞−1(1x−i−1x+i)cos(tx+t)dx=12i∫∞−1−icos(tx+t+it)xdx−12i∫∞−1+icos(tx+t−it)xdx=12icos(t+it)∫∞−1−icos(tx)xdx−12isin(t+it)∫∞−1−isin(tx)xdx−12icos(t−it)∫∞−1+icos(tx)xdx+12isin(t−it)∫∞−1+isin(tx)xdx=−12icos(t+it)Ci(−t−it)−12isin(t+it)(π2−Si(−t−it))=+12icos(t−it)Ci(−t+it)+12isin(t−it)(π2−Si(−t+it))
This matches what Mathematica computes.
If the integral was over the entire real line, the answer would avoid Ci and Si. In fact, using contour integration, we get
∫∞−∞cos(tx)x2−2x+2dx=∫∞−∞cos(tx+t)x2+1dx=cos(t)∫∞−∞cos(tx)x2+1dx−sin(t)∫∞−∞sin(tx)x2+1dx=cos(t)Re(∫∞−∞eitxx2+1dx)−0=cos(t)Re(12i∫γ(1x−i−1x+i)eitxdx)=cos(t)Re(12i∫γeitxx−idx−12i∫γeitxx+idx)=cos(t)Re(12i2πie−t−0)=πcos(t)e−t
Where γ is the contour along the real axis and circling back counter-clockwise around the upper half-plane.
This may have been the intended question.
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