In the previous two lectures, we have discussed
the properties of Fourier transform, Fourier
sine transform and Fourier cosine transform.
And also we have discussed the application
of these properties, that if I know the Fourier
transform of a function, using the properties
of Fourier transform, I can find out the Fourier
transform of other complicated properties.
So, let us discuss some more properties and
the applications of those properties, which
we have not done.
The first one is Fourier transform of x to
the power n f x, this is equals minus i to
the power n d n d alpha n of F of alpha. Similarly,
Fourier sine transform of x f x, this is equals
minus d d alpha of Fourier cosine transform
of alpha. You see here, in case of Fourier
sine transform, it is becoming minus d d alpha
of Fourier cosine transform of alpha and Fourier
cosine transform of x f x, this is equals
d d alpha of Fourier sine transform of alpha.
Let us see the proof of the first one over
here. Number 1, your F of alpha, we know it
from the definition, 1 by root 2 pi minus
infinity to infinity f x into e power i alpha
x into d x. Therefore, from here, if I have
to do d d n of F alpha, d d n d alpha n of
F alpha, this equals you can write down 1
by root 2 pi minus infinity to infinity, this
will be equals to f x into f x into not e
power. So, each time you are making the derivative.
So, at the nth time I am writing it directly
it will be i x whole to the power n into e
power i alpha x into d x.
Or in other sense, this term I have written
directly otherwise I have to show what is
d d x of d d alpha of F alpha. From there,
I have to show d square d alpha square. Since
this type of thing we have done for the case
of Laplace transform, so, directly I am writing,
because internally you are taking the derivative
inside differentiation under the sign of integration
cases. So, this I am writing f x after nth
derivative will be i x whole to the power
n into e power i alpha x into d x. So, i power
n will go outside.
So, it will be i power n into root over 2
by pi minus infinity to infinity f x into
e power f x into this will be x to the power
n will be there into e power i alpha x into
d x. And this is equals to i power n into
what is this, 1 by root 2 pi minus infinity
to infinity f x into x to the power n into
i alpha x. So, what is your function? If I
assume this, that this is my function, say
new function, say g x, which is equals f x
into x to the power n, then I can tell 1 by
root 2 pi minus infinity to infinity this
g x into e power i alpha x d x is nothing
but the Fourier transform of g x.
So, this I can write down, Fourier transform
of f x into x to the power n. So, I think
it is clear, Fourier transform of f x into
x to the power n, this is equals to 1 by root
2 pi minus infinity to infinity f x into x
to the power n into e power i alpha x into
d alpha. Therefore, Fourier transform of f
x x to the into x to the power n, this is
equals 1 by i to the power n into d n, from
here, d n d alpha n of F of alpha.
And this I can write down i to the power n.
So, this is minus i, I can make it; so, minus
i to the power n d n into d alpha n of F of
alpha. So, basically I am getting the proof
of this one that Fourier transform of f x
into x to the power n. This 1 by i to the
power n always I can write down, minus 1 whole
to the power n into d n d d d n d alpha n
of F of alpha. So, let us see quickly. These
three things we have to prove.
We are proving the first one. I am starting
with F alpha equals this and d n d alpha n
of F alpha if I make the derivative of this
function n times. Then in the integral, I
will obtain i x to the power n into e power
i alpha x. This we can show it by making first
d d alpha of F alpha is what, then d square
d alpha square, so that thing I have skipped,
because we have shown it in the case of Laplace
transform.
So, i power n I am bringing outside. And once
I am i power n I am bringing outside, 1 by
root 2 pi minus infinity to infinity, this
one is nothing but Fourier transform of f
x into x to the power n. So, once I am getting
this one, Fourier transform of x to the power
n F alpha, then I can write down Fourier transform
of x to the power n f x equals 1 by i to the
power n d n d alpha n of F alpha. And this
I can write it as minus i to the power n d
n d alpha n of F alpha. This completes the
proof of the first one.
Let us see the proof of the second one now.
I am starting with Fourier cosine transform
of alpha, this we know that root over 2 by
pi 0 to infinity f x into cos of alpha x into
d x. This we know from the definition that
Fourier cosine transform of alpha is root
over 2 by pi 0 to infinity f x cos alpha x
d x. And once I have written these, therefore,
what is your d d alpha of Fourier cosine transform
of alpha. d d alpha of Fourier cosine transform
of alpha, this is nothing but root over 2
by pi 0 to infinity, your d d alpha is there
that is f x into d d alpha of cos of alpha
x, I can write down. I can bring this under
integration inside the integral sign, so that
if I integrate this, if I differentiate this
cos of alpha x is there. So, minus will come,
x f x into sine of alpha x d x.
And what is this? This part is nothing but,
this integral is nothing but the Fourier sine
transform of x f x root over 2 by pi 0 to
infinity x f x into sine of alpha x d x is
the Fourier sine transform of Fourier x f
x, so that we can write it as minus Fourier
sine transform of x into f x. Therefore, I
can write down Fourier sine transform of x
f x, this is equals to minus this is equals
minus d d alpha of Fourier cosine transform
of alpha. So, this completes the proof that
Fourier sine transform of x f x equals minus
d d alpha of Fourier cosine transform of alpha.
In the same fashion, I can prove that Fourier
cosine transform of x f x this will be equals
to minus will not be here, d d alpha of Fourier
sine transform of alpha this we can write
down. So, therefore, this we have proved.
And on the same fashion, we can prove this
part also which you can check it of your own.
So, let us again go back to this, this already
we have shown. For the proof of the second
part, we starting with Fourier cosine transform
of alpha equals this integral from definition.
From here, we are writing d d alpha of Fourier
cosine transform of alpha as this and this
is equals minus root over 2 by pi 0 to infinity
x f x sin alpha x which is nothing but again
Fourier sine transform of x f x. So, I am
just writing minus Fourier sine transform
of x f x. So, we can tell Fourier sine transform
of x f x equals minus d d alpha of Fourier
cosine transform of alpha. And similarly as
I told, the third proof you should do it of
your own.
Now, let us see the next theorems. Number
(i), Fourier transform of f dash x that is
derivative of a function, if I know Fourier
transform of a function, then the derivative
of the function, Fourier transform of the
derivative of the function can be expressed
in terms of Fourier transform of that function.
Again this is similar properties we have studied
whenever we were doing the Laplace transform.
So, the first one is Fourier transform of
f dash x is equals to minus i alpha into Fourier
transform of x if f x approaches 0 as x approaches
plus minus infinity. Now, this particular
condition is required whenever we will prove
it, we will tell why it is required this condition
that as x approaches plus minus infinity,
f x approaches 0.
Similarly, number (ii), Fourier sine transform
of f dash x equals minus alpha into Fourier
cosine transform of alpha, where f x approaches
0 as x approaches infinity. This is obvious
because for Fourier transform, the interval
is minus infinity to infinity, whereas for
Fourier sine and Fourier cosine transform,
x approaches 0 to infinity. Similarly, for
Fourier cosine transform of f dash x, this
will be equals to alpha Fourier sine transform
of alpha minus root over 2 by pi f of 0 when
f x approaches 0 as x approaches infinity.
And the double derivative also we have shown,
Fourier cosine transform of f double dash
x equals that quantity. Fourier sine transform
of f double dash x is equals to the quantity
of this one minus alpha square into Fourier
sine transform of alpha plus alpha root over
2 by pi into f of 0. Here, only one thing
is there, in the first three cases, only f
x approaches 0. So, whenever I am taking the
second derivative, then not only f x, but
f x and its first derivative, but that is
f dash x both should approach 0 whenever x
approaches infinity. So, let us see the theorems,
proofs of the theorems one after another.
So, the first we have to prove that Fourier
transform of f dash x equals minus i alpha
into F of alpha whenever f x approaches 0
as x approaches plus infinity or minus infinity.
I am starting with the left hand side, that
is Fourier transform of f dash x equals 1
by root 2 pi minus infinity to infinity e
power i alpha x into f dash x d x. This is
equals 1 by root 2 pi minus infinity to infinity
e power i alpha x into f dash x, you can write
down d of f x from d of f x into d x, so that
now I can take it one as the first function,
other one as the second function. And I can
use integration by part, and I can integrate
it by integration by parts.
So, by integrating using integration by parts,
the first one will be e power i alpha x into
d of f x was there. So, it will be simply
f x, where x will vary from minus infinity
to plus infinity, minus if I integrate this
i alpha will come out, i alpha minus infinity
to infinity into f x into e power i alpha
x into d x. And what I know, I know that f
x approaches 0 as x approaches plus infinity
or minus infinity, this I know it, so that
this value will vanish.
Now, you can understand that I need the value
of f x at this point and if f x is 0 at x
equals plus infinity or minus infinity, then
the first term will vanish, and the second
term will be i alpha into this is nothing
but 1 by root 2 pi minus infinity to infinity
f x e power i alpha x d x is nothing but Fourier
transform of x f x and which I am writing
F of alpha. So, F of alpha is nothing but
Fourier transform of f x, which is the quantity
1 by root 2 pi minus infinity to infinity
f x into e power i alpha x into d x.
And please note one thing here. So, this completes
the proof that Fourier transform of f dash
x, this is equals to minus F of alpha minus
i alpha into F of alpha. I am not giving the
proof, but please note down this thing, Fourier
transform of f n x, nth derivative of the
function. If I know the Fourier transform
of f x, then Fourier transform of nth derivative,
I can show it simply minus i alpha whole to
the power n into F of alpha, obviously, if
f, f dash, f double dash, like this way, f
n minus 1 approaches 0 as x approaches plus
minus infinity.
In the same way, by induction I can prove
this one, you can try it of your own that
Fourier transform of f n x equals minus i
alpha whole to the power n into F of alpha,
where the condition is, please note this one
that f, f dash, f double dash up to f n minus
1 x approaches 0 whenever x approaches plus
infinity or minus infinity.
So, again let us see it quickly, these things
we had to prove. And for the first proof,
I am starting with Fourier transform of f
dash x. In the integrand, f dash x I am writing
as d of f x, so that I can use one as the
first function and other one as the second
function. And I will use integration by parts.
Using integration by parts, I am getting these
two things. The first integral, the limiting
value of the first integrand will be will
be 0, because f x approaches 0 whenever x
approaches plus infinity or minus infinity,
it is known to us, so that is 0. And second
integral is nothing but the minus i alpha
into Fourier transform of F alpha, f x.
So, Fourier transform of f x, which we are
writing as capital F alpha from the definition,
we can write down. Therefore, this completes
the proof that Fourier transform of f dash
x equals minus i alpha into F of alpha. And
on the same way, we can prove that Fourier
transform of f n x, this is equals minus i
alpha whole to the power n into F of alpha,
whenever f, f dash, f double dash, f n minus
1 x approaches 0 as x approaches plus or minus
infinity.
Now, let us see the second one. Second one,
Fourier sine transform of f dash x. On the
same way, we can approach. Root over 2 by
pi into 0 to infinity 0 to infinity f dash
x into sine of alpha x d x. And I can write
it as root over 2 by pi 0 to infinity f dash
x, I will keep it as it is. Sorry, sine of
alpha x I am writing, and f dash x is d of
f x into d x. Use the integration by parts
now. Root over 2 by pi into f x sin alpha
x 0 to infinity minus alpha into 0 to infinity
f x cos of alpha x into d x. So, the first
part will vanish, this part will be equals
to 0, because f x approaches 0 as x approaches
infinity and at 0, automatically it will be
0. So, this first value is 0, so that I can
write down minus alpha into root over 2 by
pi 0 to infinity f x into cos of alpha x d
x. And this integral is known to us, it is
nothing but Fourier cosine transform of the
function f x, which we are writing as F c
of alpha.
So, this completes the proof that Fourier
sine transform of f dash x, this is equals
minus alpha into Fourier cosine transform
of alpha. On the same way, the third one also
you can prove it, this proof is Fourier sine
transform of f dash x equals minus alpha into
Fourier cosine transform of alpha. In the
same way, you can prove Fourier cosine transform
of f dash x, this is equals my alpha into
Fourier sine transform of alpha minus root
over 2 pi 2 by pi f of 0. Because I, I do
not know the value of f 0. So, in the same
fashion whatever way I have shown, you can
prove it of your own. And you should do it
of your own. Of course, here again your f
x approaches 0 as x approaches infinity and
for that reason, here also this value was
0.
So, again let us see quickly the second one
that is Fourier sine transform of f dash x
I am finding out. So, this is root over 2
by pi sin alpha x into d of f x. This I can
break it into using integration by parts,
the limiting value of the first term will
be 0, because f x approaches 0, as x approaches
infinity. And second integral is nothing but
Fourier cosine transform of the function f
x. And so this integral, first term vanishes
and second integral is nothing but this one
so minus alpha into Fourier cosine transform
of alpha, which completes the proof that Fourier
sine transform of f dash x equals this quantity.
On the same way, the third one, we can prove
it, you can prove it of your own, which is
Fourier cosine transform of f dash x, you
can find that find out like that way.
Let us see the proof of the fourth one that
is Fourier cosine transform of f double dash
x. This is equals root over 2 by pi 0 to infinity
f double dash x into cos of alpha x d x, from
the definition we can tell this one. So, this
equals root over 2 by pi 0 to infinity cos
of alpha x into d of f dash x into d x. Now,
use again integration by parts over here.
So, you can write down root over 2 by pi f
dash x into cos of alpha x, x varies from
0 to infinity plus alpha into 0 to infinity
f dash x into cos was there. So, it will be
sine of alpha x into d x.
So, therefore, this equals again the first
part will vanish, because we know this thing
not vanish fully vanish. At infinity this
will vanish, because f x and f dash x approaches
0 as x approaches infinity, but at x equals
0, you will get a term.
So, this is sin cos 0 is 1. So, you will get
f dash 0. So, basically minus root over 2
by pi. At 0, it is negative. So, minus root
over 2 by pi f dash 0 plus alpha into, this
is nothing but again root over 2 by pi into
0 to infinity f dash x sin alpha x d x is
nothing but Fourier sine transform of f dash
x. So that I can write down this is nothing
but Fourier sine transform of f dash x.
So, therefore, this equals minus root over
2 by pi f dash 0 plus alpha into Fourier sine
transform of f dash x using the earlier property,
I can write down minus alpha into Fourier
cosine transform of alpha. This, the property
we have proved just now. So, this is equals
minus root over 2 by pi f dash 0 plus alpha
into minus this value is equals to this thing.
So, this if I write down minus this will be
equals to minus alpha square into Fourier
cosine transform of alpha minus root over
2 by pi f dash 0. So, this way I can tell
that Fourier cosine transform of f double
dash x equals minus alpha square into Fourier
cosine transform of alpha minus root over
2 by pi f dash 0.
And similarly you can prove the 5th one also
that is Fourier sine transform of f double
dash x, this is equals to minus alpha square
into Fourier sine transform of alpha plus
alpha into root over 2 by pi into f 0. Here
again f x and f dash x both approaches 0 as
x approaches infinity. So, please note this
thing, that number (v), we can prove it, you
can prove it of your own, on the same fashion
as we have proved the fourth one. So, let
us see the fourth one, Fourier cosine of f
double dash x I am finding out which is root
over 2 by pi 0 to infinity f double dash x
into cos of alpha x d x. And this I am writing
d of f dash x. Now, I am integrating by parts.
We have already shown. So, we are getting
minus root over 2 by pi alpha Fourier sine
transform of f dash x. And using the earlier
property, Fourier sine transform of f dash
x is nothing but minus alpha into Fourier
cosine transform of alpha, so that this equals
I can write down Fourier cosine transform
of f double dash x is equals to minus alpha
square into Fourier cosine transform of alpha
minus root over 2 by pi into f dash 0. This
completes the proof of the third one fourth
one. And on the same fashion, we can prove
the fifth one also as I have mentioned earlier
which you should do it of your own.
The next theorem, the earlier theorem was
on the differentiation of the Fourier series
of a function. If I know the Fourier, if I
know the Fourier transform of a function,
then what should be the, what should be the
Fourier transform of the differentiation of
the functions, that we were studying. Now,
we will start that if I know the Fourier transform
of a function, if I integrate the function,
then what should be the Fourier transform
of that function. Here, we are saying Fourier
transform of a to x f x d x is equals to F
of alpha by minus i alpha. So, let us see
the proof of this one, Fourier transform of
a to x f x d x equals F of alpha by minus
i alpha.
I am assuming this thing. Let phi x equals
a to x f x d x, phi x equals a to x f x d
x. So that this implies phi dash x is nothing
but it will be if you use differentiation
under this, you will obtain phi dash x this
is equals to f x. So that Fourier transform
of phi dash x using the earlier properties,
I can write down minus i alpha into Fourier
transform of phi x. This is equals again minus
i alpha into Fourier transform of phi x is
nothing but a to x phi x d x, this sorry a
to x not phi x, but this is f x d x, phi x
is a to x f x d x. Therefore, from here, I
can write down Fourier transform of a to x
f x d x. This is nothing but minus i alpha
will go on that side, so 1 by minus i alpha
into Fourier transform of phi dash x.
And Fourier transform of phi dash x is nothing
but minus i alpha into Fourier transform of
this phi dash x is f x, so my 1 by minus i
alpha into Fourier transform of f x. And this
equals I can write down Fourier transform
of f x is nothing but F alpha by minus i alpha
which completes the proof that Fourier transform
of a to x f x d x, this is equals to the Fourier
transform of f x divided by minus i alpha.
Let us see it quickly.
So, I have to prove this that Fourier transform
of a to x f x d x equals F alpha by minus
i alpha. So, I am assuming phi x is this,
so that your phi dash x is f x. And now I
am taking Fourier transform of phi dash x
equals using the property minus i alpha into
Fourier transform of phi x which is equals
to minus i alpha into Fourier transform of
phi x is a to x f x d x.
And so I can tell Fourier transform of a to
x f x d x minus i alpha by Fourier transform
of phi dash x. Again phi dash x is f x. So,
I am replacing it and I got the result that
Fourier transform of a to x f x d x equals
Fourier transform of f x that is capital F
of alpha minus i alpha. So, the if I know
the Fourier transform of a function and if
I integrate that function from some range
a to x f x d x, then what will be the Fourier
transform of that integration of the function,
that I can find out from this particular theorem.
