7
Fourier Integral Transforms
7.1 INTRODUCTION
Integral transforms are useful in solving initial and boundary value problems and in evaluating certain integrals. Laplace and Fourier transforms are two important transforms which are widely used in engineering and physical applications. They are used in the solution of conduction of heat, vibration of strings, oscillations of elastic beams, transmission lines, etc.
Here we define Fourier transform together with Fourier sine and cosine transforms, their inverses and study their properties and consider evaluation of certain integrals.
7.2 INTEGRAL TRANSFORMS
A linear integral transform or simply an integral transform of a function f(x) is defined by
where K(s, x), called the Kernel of the transformation, is a known function. For different choices of the Kernel K(s, x) and the limits a and b, we obtain different transforms.
7.2.1 Laplace Transform
When K(s, x) = e−^{sx} (s > 0), a = 0 and b = ∞, we obtain the Laplace transform of f(x). It is defined by
7.2.2 Fourier Transform
When K(s, x) = e^{isx}, a = −∞ and b = ∞, we obtain the Fourier transform of f(x). It is defined by
7.3 FOURIER INTEGRAL THEOREM
Let f(x) be a function satisfying the following Dirichlet conditions in every interval (−l, l) (l > 0) however large:
 f(x) is periodic; f(x) and its integrals are singlevalued and finite (bounded).
 f(x) has at most a finite number of discontinuities in any one period.
 f(x) has at most a finite number of maxima or minima.
Then the Fourier series expansion of f(x) in (−l, l) is given by
where
Substituting for a_{0}, a_{n}, b_{n} from (7.5), Eq. (7.4) becomes
If we further assume that f(x) is absolutely integrable for all x
i.e.,
Now (7.6) becomes
Putting or we have, on taking limit as n → ∞ or δs → 0
which is called the Fourier Integral (FI) of f(x).
Note 7.3.1 A rigorous proof of the above theorem is beyond the scope of this book.
Note 7.3.2 Fourier integral representation for a function f(x) is useful in solving certain differential and integral equations.
7.3.1 Fourier Sine and Cosine Integrals (FSI's and FCI's)
Writing the expansion for cosine in (7.8) we have
When f(t) is an odd function in (−∞, ∞) then the integral in the first brackets in (7.9) is zero since the integrand cos st f (t) is an odd function in (−∞, ∞) and so we obtain
which is called the Fourier Sine Integral (FSI) of f(x).
When f(t) is an even function in (−∞, ∞) then the integral in the second brackets in (7.9) is zero since the integrand sin st f(t) is an odd function in (−∞, ∞) and so we obtain
which is called the Fourier Cosine Integral (FCI) of f(x).
7.4 FOURIER INTEGRAL IN COMPLEX FORM
cos s(t − x) is an even function of s in (−∞, ∞) and so we can write (7.8) in the form
Also, sin s(t − x) is an odd function of s in (−∞, ∞). So,we can write
Now, multiplying (7.13) by ‘i’ and adding it to (7.12) we obtain
Using Euler's formula e^{iθ} = cos θ + i sin α we can write the above integral as
which is called Fourier Integral of f(x) in complex form (FICF)
7.4.1 Fourier Integral Representation of a Function
By (7.9) a function f(x) may be represented by a FI as
and
If f(x) is an odd function in (−∞, ∞) then
where
and if f(x) is an even function in (−∞, ∞) then
where
7.5 FOURIER TRANSFORM OF f(x)
Writing the exponential function e^{ix(t−x)} as a product of two exponential functions e^{ix(t−x)} = e^{ist} · e^{−isx} the FI (7.14) may be put in the following form
The expression inside the brackets is a function of s. Denoting it by f(s) and replacing t by x, we have the Fourier transform (FT) of f(x). Then the inverse Fourier transform (IFT) of F(s) is given by (7.22).
Fourier Transform
The Fourier transformation of f(x), denoted by F{f(x)}, is defined by
Inverse Fourier Transform
The inverse FT of F(s), denoted by F^{−1} {F(s)} is defined by
Thus, the function F(s) defined by (7.23) is called the Fourier Transform of f(x) and is denoted by
and the function f(x) given by (7.24) is called the Inverse Fourier transform of F(s) and is denoted by
The process of obtaining the FT F{f(x)} = F(s) from a given function f(x) is called the FT method or simply Fourier Transform.
Existence of Fourier Transform
The following conditions are sufficient for the existence of the FT of a function f(x):
 f(x) is piecewise continuous on every finite interval.
 f(x) is absolutely integrable for all x.
7.5.1 Fourier Sine Transform (FST) and Fourier Cosine Transform (FCT)
The FSI in (7.10) can be written by replacing t by x as
The expression in the brackets in (7.27) is a function of s denoted by F_{s}(s). Writing
Equation (7.20) becomes
The function defined by (7.28) is called the Fourier sine transform (FST) of f(x) and that defined by (7.29) is called the inverse Fourier sine transform of F_{s}(s).
The Fourier cosine integral (7.11) can be written by replacing t by x as
The expression in the brackets (7.30) is a function of s denoted by F_{c}(s). Writing
The functions F_{c}(s) and F_{c}^{−1} {F_{c}(s)} = f(x) are, respectively, called the Fourier cosine transform and inverse Fourier cosine transform, respectively.
7.6 FINITE FOURIER SINE TRANSFORM AND FINITE FOURIER COSINE TRANSFORM (FFCT)
The finite Fourier sine transform of f(x) in 0 < x < l is defined by
where n is an integer.
The function f(x) is then called the inverse finite Fourier sine transform of F_{s}(n) and is given by
Similarly the finite Fourier cosine transform of f(x) in 0 < x < l is defined by
where n is an integer.
The function f(x) is then called the inverse finite Fourier cosine transform of F_{c}(n) and is given by
Note 7.6.1 The finite Fourier sine transform is useful for solving problems with boundary condition of heat distribution (say) on two parallel boundaries while the finite Fourier cosine transform is useful for solving problems in which the velocity distribution (say) normal to two parallel boundaries are prescribed.
7.6.1 FT, FST and FCT Alternative definitions
The Fourier transform, FST and FCT are alternatively defined as follows:
Definition 7.6.2
Definition 7.6.3
Note 7.6.4 Each of these notations has its advantages and disadvantages. So, one can adopt any one of these definitions. If both transformation and inverse transformation are involved completing one cycle there will not be any difference in the results. If only one of these is involved the results will be different. The student, while answering a question, should understand which notation is used and then answer, clearly stating the notation he or she is using.
7.7 CONVOLUTION THEOREM FOR FOURIER TRANSFORMS
7.7.1 Convolution
Definition 7.6.5 The convolution of two functions f(x) and g(x) over the interval (−∞, ∞) denoted by f * g is defined by
7.7.2 Convolution Theorem
Theorem 7.6.6 If F(s) = F{f(x)} and G(s) = F{g(x)} are the Fourier transforms of f(x) and g(x) then the Fourier transform of the convolution of f(x) and g(x) is the product of their transforms
Proof By the definition of Fourier transforms, we have
7.7.3 Relation between Laplace and Fourier Transforms
Proof
Thus, the Fourier transform of a function f(t) defined by (7.45) is the Laplace transformation of g(t).
7.8 PROPERTIES OF FOURIER TRANSFORM
7.8.1 Linearity Property
Theorem 7.8.1 If F(s) and G(s) are the Fourier transforms of f(x) and g(x), respectively, then
where a and b are any constants.
Proof By definition we have
7.8.2 Change of Scale Property or Damping Rule
Theorem 7.8.2 If F(s) is the complex Fourier transform of f(x) then
Proof Since F(s) is the complex Fourier transform of f(x) we have
So, the complex Fourier transform of f(ax) is given by
Corollary 7.8.3 If F_{s}(s) and F_{c}(s) are the Fourier sine transform and Fourier cosine transform of f(x), respectively, then F_{s}(f(ax)) = (1/a)F_{s}(s/a) and F_{c}(f(ax)) = (1/a)F_{c}(s/a).
7.8.3 Shifting Property
Theorem 7.8.4 If F(s) is the complex Fourier transform of f(x) then
Proof It is given that
By definition
where u = x − a and du = dx.
7.8.4 Modulation Theorem
Theorem 7.8.5 If F(s) is the complex Fourier transform of f(x) then
Proof It is given that
 By definition
 can be proved similarly.
Corollary 7.8.6 If F_{s}(s) and F_{c}(s) are the Fourier sine transform and the Fourier cosine transform of f(x), respectively, then
Example 7.8.7 Using Fourier integral show that
Solution The Fourier cosine integral is given by
Let f(t) = e^{−at} so that f(x) = e^{−ax}. Substituting in the above integral, we have
Example 7.8.8 Prove that
Solution Putting a = 1 in Example 7.1, we obtain
Example 7.8.9 Express as a Fourier integral and hence evaluate
Solution The Fourier integral of f(x) is given by
The funtion f(t) is defined by
∴ We obtain
which is the Fourier integral of f(x). From (7.55) we obtain
At x = ±1, f(x) is discontinuous and the value of the integral is
Example 7.8.10 Prove that
Solution From Example 7.3
Putting x = 0 we have
where we have replaced the dummy variable s by x.
Example 7.8.11 Using the Fourier integral representation, show that
Solution Let f(x) be defined by
We now find the Fourier integral representation of f(x) in the exponential form, which is given by
The function f(x) is discontinuous at x = 0 and its value is
Example 7.8.12 Express as a Fourier integral. Evaluate

[JNTU 2003(2), 2004(3)]
Solution The Fourier integral of f(x) is
For which are points of discontinuity the value of the integral is
Choosing x = 0 and a = 1 we obtain where we have replaced the dummy variable s by x.
Example 7.8.13 Express the function as a Fourier sine integral.
Hence, evaluate
Solution The Fourier sine integral of f(x) is given by
At x = π which is a point of discontinuity for f(x) the value of the integral is
Example 7.8.14 Find the Fourier transform of
Hence, evaluate
[JNTU 2002, 2004S]
Solution The Fourier transform of f(x) is given by
By the inverse Fourier transform, we have
Putting x = 1/2 we obtain from the above equation (7.62)
Equating the real parts on either side
since the integrand is even in (−∞, ∞) and we have replaced the dummy variable s by x.
Example 7.8.15 Find the Fourier transform of
Hence, evaluate
Solution The Fourier transform of f(x) is given by
We have and hence F{f(x)} = F(s) = 2 when s = 0 (7.66)
By the inverse Fourier transform we have
Putting x = 0 we get
Since the integrand is an even function in (−∞, ∞) we have
where we have replaced the dummy variables s by x.
Example 7.8.16 Find the Fourier (a) cosine and (b) sine transform of f(x) = e^{−ax} (x > 0, a > 0).
Deduce the values of
Solution
 The Fourier cosine transform of f(x) is given by
The inverse Fourier cosine transform of F_{c}(s) is
 The Fourier sine transform of f(x) is given by
The inverse Fourier sine transform of F_{s}(s) is
Corollary 7.8.17 Putting a = 0 in the above result we obtain
Example 7.8.18 Find the Fourier sine transform of Hence, show that
Solution The variable x is positive in (0, ∞) so that for (0, ∞).
The Fourier sine transform of is given by
By the inverse Fourier sine transform of F_{s}(s) we have
If we replace ‘x’ by ‘a’ we obtain
where we have replaced the dummy variable s by x.
Hence
Example 7.8.19 Find the Fourier sine transform of e^{−ax/x}.
Solution By definition, the Fourier sine transform of f(x) is
Differentiating both sides with respect to ‘s’
Integrating now with respect to ‘s’ we obtain
Example 7.8.20 Find the Fourier cosine transform of
[JNTU2002, 2003S, 2004S]
Solution The Fourier cosine transform of f(x) is given by
Example 7.8.21 Find the Fourier cosine transform of f(x) = 1/(1 + x^{2}). Hence, derive the Fourier sine transform of
Solution By definition, the Fourier cosine transform of f(x) is given by
Differentiating both sides of (7.87) with respect to ‘s’
Differentiating both sides of (7.88) with respect to s again
The general solution of (7.89) is
When s = 0, (7.87) and (7.90) give
Also
Solving (7.92) and (7.93) we get
From (7.87) and (7.90) we have
Now
Example 7.8.22 Find the Fourier sine and cosine transform of e^{−ax}(a > 0) and deduce the inverse formula.
[JNTU 2002, 2004]
Solution The Fourier sine transform of f(x) = e−^{ax} is
The inverse Fourier sine transform for is
The Fourier cosine transform of f(x) = e^{−ax} is
is
Example 7.8.23 Solve the intergral equation
Hence evaluate
Solution By definition, the Fourier cosine transform of f(x) is
By the inversion formula we have
Now
From (7.101) and (7.102) we obtain
where we have replaced the dummy variable t by x.
Example 7.8.24 Show that the Fourier transform of is reciprocal.
[JNTU 2002, 2004S]
Solution Here we have to use definition 2 for the Fourier transform, viz.
Hence
Here the function f(x) and the Fourier transform F(s) are the same. Hence, is selfreciprocal.
Note 7.8.25 If we use the other definitions there will be difference of a constant factor.
Example 7.8.26 Find the Fourier sine transform of (l/x)(x^{2} + a^{2}).
Hence, deduce the Fourier cosine transform of
[JNTU 2001, 2002, 2003]
Solution The Fourier sine transform of f(x) is given by
Now, let
Differentiating both sides of (7.105) with respect to ^{‘}s^{’}
Again differentiating both sides of (7.106) with respect to ^{‘}s^{’}
which is a secondorder linear differentialequation (7.107)
The general solution of (7.107) comprises two parts: the complementary function (CF) and the particular intergal (PI).
∴ The general solution is
Putting s = 0 in (7.105) and (7.108) we obtain
and similarly from (7.106) and (7.109) we obtain
Solving (7.110) and (7.111) we obtain
Substituting these values in (7.108) we have
which is the required Fourier sine transform of
Now we deduce the Fourier cosine transform of
The Fourier cosine transform of is given by
∴ The Fourier cosine transform of g(x) = 1/(x^{2} + a^{2}) is G_{c}(s) = (π/2a)e^{−as}. (7.114)
EXERCISE 7.1
 Find the Fourier transform of e^{−x2}, (−∞ < x < ∞).
Ans:
 Find the Fourier transform of
Ans:
 [(a^{2}s^{2} − 2)sinax + 2ascosas]
 Express as a Fourier integral
Hence, evaluate .
Ans:
 Using Fourier integral show that
 Using Fourier integral show that
 Using Fourier intergral representation show that
 Prove that
[Hint: Use FSI with f(t) = e^{−at} − e^{−bt}.]
 Find the Fourier sine transform of
Ans:
 Find the Fourier sine transform of
Ans:
 Find the Finite fourier sine and cosine transforms of f(x) = 2x, 0 < x < 4.
Ans:
 Solve
Ans:
 Solve
Ans: f(x) = (2 + 2 cos x − 4 cos 2x)/πx
 Find the Fourier cosine transform of
 e^{−ax}cos ax
 e^{−ax}sin ax.
Ans:
 Find the Fourier cosine transform of .
Ans:
7.9 PARSEVAL'S IDENTITY FOR FOURIER TRANSFORMS
Theorem 7.9.1 If F(s) and G(s) are the Fourier transforms of f(x) and g(x), respectively, then
where denotes the complex conjugate of g.
Proof Let g(x) be the inverse Fourier transform of G(s) so that
 Taking conjugates on both sides
(Changing the order of integration)
by the definition of Fourier transform
 Setting g(x) = f(x) for all x we obtain
7.10 PARSEVAL'S IDENTITIES FOR FOURIER SINE AND COSINE TRANSFORMS
Theorem 7.10.1 Similarly, we can obtain the following results:
Example 7.10.2 If find the Fourier transform.
Using Parseval's identity prove that
Solution The Fourier transform of f(x) is given by
Parsevel's identity for Fourier transform is
Example 7.10.3 Using Parseval's identities prove that
Solution The Fourier sine transform of is given by
By Parseval's identity for Fourier sine transform we have
where we have replaced the dummy variable s by x.
Example 7.10.4 Evaluate using Parseval's identity.
Solution The Fourier cosine transform of f(x) = e^{−ax} is given by
and the Fourier cosine transform of g(x) = e^{−bx} is given by
Parseval's identity for Fourier cosine transforms is
Example 7.10.5 Evaluate
using Parseval's identity.
Solution The Fourier cosine transform of f(x) = e^{−ax} is given by
EXERCISE 7.2
Using Parseval's identities prove the following: