This document discusses Fourier integrals, which extend Fourier series to non-periodic functions. It defines the Fourier cosine integral and Fourier sine integral. As an example, it finds the Fourier cosine and sine integrals of the function f(x)=e-kx for x>0 and k>0. The Fourier cosine integral of this function is 2k/π∫0∞ cos(ωx)/(k2+ω2) dω and the Fourier sine integral is 2/π∫0∞ ωsin(ωx)/(k2+ω2) dω. In conclusion, Fourier integrals are used to find the Fourier representation of functions defined on the whole x-axis.

1. Study of Fourier
intigrals

2. -:CONTENT:-
Fourier integrals
Fourier cosine integral
Fourier sine integral
Problem
Conclusion

3. -: Fourier integrals :-
Fourier integral is an extension of Fourier series in non-periodic functions. Hear integration is used
instead of Summation in a Fourier series.
The Fourier integrals of a function f(x) is given by.
𝑓 𝑥 =
0
∞
𝐴 𝜔 cos 𝜔𝑥 + 𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥 𝑑𝜔
Where,
𝐴 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑢 cos 𝜔𝑢 𝑑𝑢
𝐵 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑢 𝑠𝑖𝑛𝜔𝑢 𝑑𝑢
In this there are three type of integral.
I. Fourier integral
II. Fourier cosine integral
III. Fourier sine integral

4. -:Fourier cosine integral:-
Suppose f(x) is an even function .
As we know cos𝜔x is an even function and sin𝜔𝑥 is an odd function .There fore f(x) cos𝜔𝑥 is an
Even function & f(x) sin𝜔𝑥 is an odd function.
Now,
𝐴 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑢 𝑐𝑜𝑠𝜔𝑢 𝑑𝑢 =
2
𝜋 0
∞
𝑓 𝑢 𝑐𝑜𝑠𝜔𝑢 𝑑𝑢
𝐵 𝜔 =
1
𝜋 −∞
∞
𝑓(𝑢) sin 𝜔𝑢 𝑑𝑢 = 0
Fourier cosine integral represented by
𝑓 𝑥 =
0
∞
𝐴(𝜔) cos 𝜔𝑥 𝑑𝜔

5. -:Fourier sine integral:-
Suppose f(x) is an even function. Now sin 𝜔𝑥 is an odd function then f(x) cos𝜔𝑥 is also odd function and
f(x) sin 𝜔𝑥 is an even function.
Now,
𝐴 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑢 cos 𝜔𝑢 𝑑𝑢 = 0
𝐵 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑢 sin 𝜔𝑢 𝑑𝑢 =
2
𝜋 0
∞
𝑓 𝑢 sin 𝜔𝑢 𝑑𝑢
Fourier sine integral represented as
𝑓 𝑥 =
0
∞
𝐵(𝜔) sin 𝜔𝑥 𝑑𝜔

6. Problem:-
Find the Fourier cosine and Fourier sine integral of 𝑓 𝑥 = 𝑒−𝑘𝑥
where x>0 and k>0.
Ans:-
Fourier cosine integral of f(x) is given by
𝑓 𝑥 =
0
∞
𝐴(𝜔) cos 𝜔𝑥 𝑑𝜔 … . . (1)
Where,
𝐴 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑥 cos 𝜔𝑥 𝑑𝑥
=
1
𝜋 −∞
∞
𝑒−𝑘𝑥
cos 𝜔𝑥 𝑑𝑥
Since f(x) is even so the integration is even
=
2
𝜋 0
∞
𝑒−𝑘𝑥
cos 𝜔𝑥 𝑑𝑥
Now, by integration by parts
=
2
𝜋
−𝑘
𝑘2+𝜔2 𝑒−𝑘𝑥 −𝜔
𝑘
sin 𝜔𝑥 + cos 𝜔𝑥 ∞
0
=
2
𝜋
0 +
𝑘
𝑘2+𝜔2 =
2𝑘
𝜋 𝑘2+𝜔2
By substituting 𝐴 𝜔 into (1) we obtain the Fourier cosine integral
𝑓 𝑥 =
2𝑘
𝜋 0
∞
cos 𝜔𝑥
𝑘2 + 𝜔2
𝑑𝜔

7. Fourier sine integral of f(x) is given by
𝑓 𝑥 =
0
∞
𝐵(𝜔) sin 𝜔𝑥 𝑑𝜔 … … (2)
Where, 𝐵 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑥 sin 𝜔𝑥 𝑑𝑥
Since f(x) is odd the integral is even
=
2
𝜋 0
∞
𝑒−𝑘𝑥
sin 𝜔𝑥 𝑑𝑥
Now, by integration by parts
=
2
𝜋
−𝜔
𝑘2 + 𝜔2
𝑒−𝑘𝑥
𝑘
𝜔
sin 𝜔𝑥 + cos 𝜔𝑥
∞
0
=
2
𝜋
0 +
𝜔
𝑘2 + 𝜔2
=
2
𝜋
𝜔
𝑘2 + 𝜔2
By substituting B 𝜔 into (2) we obtain the Fourier cosine integral
𝑓 𝑥 =
2
𝜋 0
∞
𝜔 sin 𝜔𝑥
𝑘2 + 𝜔2
𝑑𝜔

8. -:Conclusion:-
Many problems involve functions that are non –periodic and are of interest on the
whole x-axis to find Fourier series of such function we use Fourier integrals.

9. THANK YOU