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3705Tut7-W13
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MATH3705 Tutorial 7
Determine the following:
1. F(e−0.5|x|).
Solution: Let f(x) = e−|x|. Then F(e−0.5|x|) = F(f(0.5x)) = 10.5
f̂(
λ0.5
)= 4
1+4λ2 .
2. F(e−|x−4|).
Solution: Let f(x) = e−|x|. Then by the second shifting theorem, F(e−|x−4|) =
F(f(x− 4)) = ei4λf̂ (λ) = 2ei4λ
1+λ2 .
3. F(e−ixe−|x|).
Solution: Let f(x) = e−|x|. Then by the first shifting theorem, F(e−ixe−|x|) = f̂ (λ− 1) =2
1+(λ−1)2.
4. F(e2ix−|x+3|).
Solution: By the second shifting theorem, F(e−|x+3|) = 2e−i3λ
1+λ2 . Then by the first shift-
ing theorem, F(e2ix−|x+3|) = F(e2ixe−|x+3|) = 2e−i3(λ+2)
1+(λ+2)2.
5. F−1{
e−i3λ
1+λ2
}.
Solution: By the second shifting theorem, F−1{
e−i3λ
1+λ2
}= 1
2e−|x+3|.
6. F−1{
11+(λ+5)2
}.
Solution: By the first shifting theorem, F−1{
11+(λ+5)2
}= 1
2e5ix−|x|.
1
7. F{xe−3x2
}.
Solution:
F(xe−3x2
)= F
(−1
6[e−3x2
]′)
= −1
6F([e−3x2
]′)
= −1
2(−iλ)F
(e−3x2
)=
iλ
2
√π√3e−λ2/12.
8. F{(x+ 2)e−(x+2)2
}.
Solution: F{(x+ 2)e−(x+2)2
}= e−2iλF
(xe−x2
)= e−2iλ i
√πλ2
e−λ2/4.
9. F{xe−x2+3ix
}.
Solution: F{xe−x2+3ix
}= F
{e3ixxe−x2
}= i
√π(λ+3)2
e−(λ+3)2/4.
10. F−1{e−λ2
}.
Solution: F−1{e−λ2
}= 1
2√πe−x2/4.
11. F−1{e2iλ−λ2
}.
Solution: F−1{e2iλ−λ2
}= 1
2√πe−(x−2)2/4.
12. F−1{λe−λ2
}.
Solution: F−1{λe−λ2
}= −ix
4√πe−x2/4.
13. F−1{e−(λ+2)2
}.
Solution: F−1{e−(λ+2)2
}= 1
2√πe2ix−x2/4.
2
14. F{xe−3x2
}.
Solution:
F(xe−3x2
)= F
(−1
6[e−3x2
]′)
= −1
6F([e−3x2
]′)
= −1
2(−iλ)F
(e−3x2
)=
iλ
2
√π√3e−λ2/12.
15. F{(x+ 2)e−(x+2)2
}.
Solution: By the second shifting theorem,
F{(x+ 2)e−(x+2)2
}= e−2iλF
(xe−x2
)= e−2iλ i
√πλ
2e−λ2/4.
16. F{xe−x2+3ix
}.
Solution: By the first shifting theorem and the result for F{xe−x2
}, we have F
{xe−x2+3ix
}=
F{e3ixxe−x2
}= i
√π(λ+3)2
e−(λ+3)2/4.
17. F−1{e−λ2
}.
Solution: Since F−1{e−tλ2
}= 1
2√tπe−x2/(4t), we have F−1
{e−λ2
}= 1
2√πe−x2/4.
18. F−1{e2iλ−λ2
}.
Solution: Note that
F−1{e−λ2
}=
1
2√πe−x2/4.
By the second shifting theorem,
F−1{e2iλ−λ2
}= F−1
{e2iλe−λ2
}=
1
2√πe−(x−2)2/4.
3
19. F−1{λe−λ2
}.
Solution:
F−1{λe−λ2
}= F−1
{d
dλ
(−1
2e−λ2
)}= ixF−1
{−1
2e−λ2
}= −ix
2
1
2√πe−x2/4 =
−ix
4√πe−x2/4.
20. F−1{e−(λ+2)2
}.
Solution: By the first shifting theorem and the question 4 above, F−1{e−(λ+2)2
}=
12√πe2ix−x2/4.
4