so welcome to the lecture series on mathematical methods and its applications so we have already
discussed fourier series we have seen what fourier series are and how we can solve some
problems based on that what are properties what are convergence properties of the fourier
series that also we have seen and also the complex of fourier series now for a integrals
what for a integrals are and why they are important let us see so if you have a periodic
function f x which is defined on a real line r then f x can be represented it by a fourier
series that we have already seen ok which is given by a naught by two plus summation
n cos n pi x pi l plus summation b n sin n pi x by l that we have already seen
if f x is not periodic suppose f x is not a periodic function then it cannot be represented
it by a fourier series over entire real line is one two express a non periodic function
by a fourier series then it is not possible however f x may be represented it in an integral
form ok how let us see now first of all suppose f x have the following three properties f
x piecewise continuous on every interval from minus l to plus l f x is absolutely integrable
on x axis that means integral from minus infinity to plus infinity mod of f x converges that
means this value is finite integral minus to plus infinity mod f x d x value is finite
and for every x belongs to r f x has left and right hand derivatives
so first of all let us suppose the function f x is periodic ok and satisfy these three
properties ok again suppose f x is a periodic function with period two will define interval
minus l two plus l satisfy the properties p one to p three which we have discussed then
it can be specified fourier series in this way ok basically we derive the fourier integral
from the fourier series itself ok we have the fourier series like this we already know
this f x suppose to a naught upon two plus summation a n cos n pi x upon l plus b n sin
n pi x by l so this is the fourier series expression and it is varying from one to infinity
or function f x ok we are a naught is given by a naught is given by one upon l integral
minus l two plus l f t d t a n is given by one by l integral minus l two l f t cos n
pi t by l into d t and b n is given by one by l integral minus l to plus l f t sin n
pi t by l into d t so these are expressions ok now now let
us substitute these expressions over here ok so what will obtain f x will be equals
to one by two l integral minus l to plus l f t d t plus summation n varying from one
to infinity a n is nothing but one by l integral minus l to plus l f t cos n pi t
by l into d t into cos n pi x upon l ok plus summation n varying from one to infinity again
one by l integral minus l to plus l f t sin n pi t by l into d t and whole multiplied
by sin n pi x by l ok now let us suppose let omega n is n pi by l ok so what will be delta
omega delta omega is omega n minus omega n minus one which is nothing but n pi by l minus
n minus one pi by l which is pi by l so let us substitute these terms in this expression
ok so what will be effects what effects we will obtain so f x will be nothing but is
equal to now now this this one by l one by l is nothing but delta omega by pi so it
is delta omega by two pi integral minus l to plus l f t d t ok from this from this expression
now plus again this one by l will be nothing but pi upon delta omega upon pi from this
expression so one upon pi will be outside one upon pi will be outside summation n varying
from one to infinity minus l to plus l f t cos and pi n pi l by l is nothing but omega
n omega n t ok into d t and whole multiplied by cos omega n into x into delta omega ok
delta omega by pi is here instead of this so delta omega is here and similarly plus
one by pi summation n varying from one to infinity integral minus l to plus l f t sin
omega n t into d t and whole multiplied by sin omega n x into delta omega because pi
n pi l by l is nothing but omega n so it is omega n x into delta omega ok
now let l now let let l tend to infinity because you on to expend the function over
the entire real line ok we want to expend the function over the entire real line thats
why we are we are tending l to infinity so as l tends to infinity of course minus l to
plus l interval will tend to minus into plus infinity that is over the entire real line
and delta omega will tend to zero delta omega is nothing but pi upon l and as l tend to
infinity this will tends to zero ok so and function f x is absolutely convergent that
we are already assumed ok so this term so this integral when l tends to infinity is
finite and delta omega tend into zero the first term is tend to zero because delta omega
tend to zero and as l tend to infinity so this value since is absolutely convergent
absolutely convergence so this value is finite finite into zero will be zero so first term
is zero and this will sum up as remaining sum because because l tends to infinity
l tends to infinity and this delta omega which tend zero so this sum up as remaining sum
so this will be nothing but one upon pi integral zero to infinity ok zero to infinity integral
minus infinity plus infinity f t cos omega t into d t whole with cos omega x into d omega
plus plus integral zero to infinity integral minus to plus infinity f t it is sin omega
t into d t and whole multiplied by sin omega x into d omega ok because l is tend to infinity
ok and delta omega tend to zero so this will be sum up as remaining sum so which is given
by one upon pi integral zero to infinity minus to plus infinity this term plus this term
so thats what we have given the p p t also that this presumed as a remaining sum of the
different integral as l tend to infinity so this will give this value
now here if you take if you take this as a omega because this is a function of omega
and this as b omega so what will be f x f x will be nothing but one by pi times integral
zero to infinity a omega cos omega x t omega plus plus integral zero to infinity it
is b omega sin omega x into d omega so that will be that will be this expression where
a omega is nothing but minus infinity to plus infinity f t cos omega t d t and b omega is
nothing but minus will plus infinity f t sin omega t into d t so this is what the fourier
integral is i mean integral representation of a function f x ok for a integral representation
of a function f x given by this where a omega is a this and b omega is this ok so we we
expend the function over the entire real line by assuming that l is tend to infinity
ok so this is how we can obtain this thing now this is called fourier integrals of
f x now suppose f x satisfy the properties p one to p three ok which we have discussed
then the fourier integral of f x converges to f x the point of continuity the same as
in fourier series this we have also discussed fourier series also the same follows in fourier
integrals also that if if satisfy the some properties p one to p three then the fourier
integral of f x converges f x at point of continuity and for discontinuity it converges
to the average of left hand limit and the right hand limit at that point ok now let
us try this problem now we want fourier integral representation of
this function so we know that the fourier integral representation of a function is given
by this expression f x is given by this expression a omega is given by this b omega is given
by this expression now their function is what function is defined function is one when mod
x is less than one and zero when mod x is greater than one ok so this is the function
now we want fourier integral representation of this function so this function is defined
over the entire real line so we want the fourier integral representation of this function how
you can find it so we first find a omega and b omega we substitute the values of a omega
b omega over here so that will give the fourier integral representation of this function so
what is a omega let us find a omega first what is a omega from here a omega is minus
plus infinity f t cos omega t d t now it is defined from minus one to plus
one ok is one so minus one to plus one is one otherwise it is zero so it is cos omega
t d t and cos omega t is an even function so it is nothing but two times zero to one
cos omega t d t and it is equals to two times sin omega t upon omega from zero to one which
is nothing but two sin omega upon omega because sin zero is zero ok now what is b omega b
omega will be nothing but again minus infinity plus infinity by this definition f t sin omega
t into d t which is equals to minus one to plus one sin omega t into d t because from
minus one to plus one it is one and it is an odd function it is zero so we have computed
both a omega and b omega and the fourier integral representation of a function f x is given
by this expression as you already know so this will be nothing but one upon pi integral
zero to infinity now what is a omega is two sin omega cos omega
x upon omega into d omega d omega is zero so that will be the fourier integral representation
of this function f x ok now we can also write this function as zero to infinity
two sin omega cos omega x upon omega d omega which will be equals to pi times f x and which
is equals to pi you multiply simply that by pi when mod x less than one that is zero when
mod x greater than one ok this two you can also bring here pi by two so it is pi by two
and what happens when x is one when x is one so it is it is discontinues so it converges
to average of left hand limit and right hand limit which is nothing but what will be f
one f one will nothing but half of f one minus plus f one plus so that will be nothing but
one by two one plus zero that is one by two and you have to multiply pi by two also so
that will be pi by four ok so in this way you can find out the value
the same you can find out when x equals to minus one also when x equal to minus one we
are the same value it is x equal to x equal plus or minus one ok so so this integral
is equal to basically this integral now we have to find out integral zero to infinity
sin omega upon omega d omega so that will be nothing but you substitute x as zero ok
its a function of omega so you can arbitrary choose any value of x so when x is zero so
when x is zero so this is nothing but so you can substitute both side as
zero so it is zero to infinity sin omega upon omega d omega is equals to pi by two f zero
and where it is zero when f is zero f is one it is pi by two ok you can directly see from
here when x is zero value is pi by two value of this expression is pi by two so you can
easily see that this value when x is zero ok in fact when you put x equal to say half
when you put x equal to half to when x equal to half satisfy this expression when x equal
to half the value of that will also be pi by two you can easily find out under different
values of x the value of this expression nothing but this expression ok now let us try to prove
this that this is equal to this expression so how we can prove this this is nothing but
fourier integral representation of function f x so what is the fourier integral representation
of function so what is the expression it is zero to i mean infinity it is cos it is
cos pi omega by two into cos omega x upon one minus omega square d omega is equals to
it is given to you pi by two cos x when mod x is less than pi by two and zero when mod
x is so we have to prove this ok we have to prove this so what is the fourier integral
representation of function f x is nothing but one upon pi times integral integral
zero to infinity a omega cos omega x t x cos omega x d omega plus integral zero to infinity
d omega sin omega x into d omega ok is it so fourier integral representation is into
d omega let us same thing and here it is d t ok now what is what is a omega a omega is
minus minus will plus infinity it is f t cos omega t d t and d omega is given by minus
will plus infinity f t sin omega t d t that you already know now if you compare this
with this so what will be a omega so it is pi f x which is equals to if you compare
this by this so cos omega x coefficient here is simply a omega
so we can simply say that a omega is nothing but cos pi omega by two upon one minus omega
square and d omega is zero because there is no term of sin omega x ok and this integral
zero to infinity a omega cos omega x d x is nothing but pi times f x which is equals to
which is equals to pi by two cos x when mod x is less than pi by two and zero when x is
mod x is greater than pi by two so this implies f x is cos x by two when mod x is less than
pi by two and zero when mod x is greater than pi by two ok so you take this f x you take
this f x and try to obtain the value of a omega if you obtain a omega as this that means
we have proved ok any how any how we have assume that this is true ok to assuming this
to be true we have find out the coefficients a omega and d omega and we have find f
x for this f x if a omega comes out to be this that means we have proved ok so what
will be a omega for this f x and b omega a omega should be this b omega should be this
so what will be a omega a omega will be given by minus infinity plus infinity f t cos omega
t d t now it is given by f x is defined like this from minus pi by two to plus pi by two
it is cos t upon two into cos omega t d t which is nothing but it is even function it
is two times by two zero to pi by two cos t into cos omega t d t which is equals to
now two two cancels out multiply divided by two again zero to pi by two it is two cos
a cos b which is nothing but sin a plus b plus sin a minus b it is equals to one by
two integral of sin is minus cos so put minus outside put minus on outside it is cos of
one plus omega t upon one plus ok both will be cos sorry cos a to cos a cos b is nothing
but cos a plus b plus cos a minus b so it will be cos it will be cos and minus is not
so it will be sin and plus sin one minus omega t upon one minus omega and t varying from
zero to pi by two so this will be nothing but one by two it is sin one plus omega into
pi by two upon one plus omega plus sin one minus omega into pi by two upon one minus
omega ok now we were simplify this is nothing but one by two
now sin ninety plus theta is sin theta to that is sin pi by two omega upon one plus
omega again sin ninety minus theta no sin ninety plus theta is cos theta sorry its cos
theta so it will be cos theta again here it is cos theta so this will comes out this cos
pi omega by two will come out and it will be nothing but two upon one minus omega square
so two two cancels out so this will comes equal to this cos pi omega by two into one
upon one minus omega square now when you compute b omega b omega must be zero when you compute
b omega for the same problem so b omega will be nothing but what minus plus infinity f
t sin omega t d t now it is nothing but minus pi two plus pi by two it is cos t by
two sin omega t d t its an odd function so it is zero so hence we have proved because
we have obtained that a omega is this and b omega is zero for this function that means
this results hold ok so we have shown the result so this how we can find out fourier
integral representation of a function f x ok so so
thank you