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