Today, we will discuss few tests, which will
decide about the convergence or divergence
of the improper integral.
We have already discussed these two examples.
The first example which 
we have discussed where . This is improper
integral of the kind one, and we say this
integral converges, if and diverges if . So,
this we have seen as an example and it will
be used for future work.
Another example which we have seen, if the
function is not defined at the end points,
that is function becomes unbounded at either
a or b, then in that case, we can also have
a convergence test and for which this example
will be helpful. So, we have seen this example
0, where the function becomes undefined is
not defined at 0. Hence this is an improper
integral of the kind two, and we have seen
this integral converges if and diverges if
. So, these two examples we have seen, and
then we wanted to generalize this particular
this result for a for an improper integral
to test whether it is convergent or not. So,
this example will be used.
So, first we will go to the convergence test
for indefinite integral sorry for infinite
integrals. So, the first test we say is a
comparison test. The comparison test, say
we know, if f(x)> 0, when , then this integral
is monotonic increasing. That means when x
increases, the corresponding value of this
integral will increase. Therefore, if we say
that this integral converges, this gives an
idea.
Therefore, we say the improper integral converges
if the integral remains bounded by k; that
is the value is always be less than k for
all values of x, where k is a fixed number,
this is a fixed real number, some positive
number. So, if this is true, then we say this
integral converges because this is a monotonic
increasing function and for each x this is
bounded by k, is less than equal to dominated
by k; k is fixed. So, the value of this integral
will be a finite quantity. So, it is convergent
and this suggests the test which is known
as the comparison test.
So, we have this test which is known as the
comparison test. What this says is if f(x)
and g(x) are positive and integrable, and
f(x) is less than equal to g(x) throughout
the interval, when . It means from a onward
whatever the point x is, this condition is
satisfied for all x; this is true for all
x; even this is true for in fact all . Then
the behavior of this integral, then we say
integral, this integral is convergent when
integral is convergent. It means when the
function f and g are two functions which are
positive and integrable, and if the right
hand side integral of g(x).
It means if we want to test the nature of
the convergence of this integral and we are
aware to construct a function g for which
this condition is satisfied throughout the
interval x greater than a, then the convergence
of this integral will imply the convergence
of this integral. Similarly, if suppose, if
, for all and if integral , then the integral
is divergent, is also divergent. So, this
is known as the comparison test. It means
with the help of the function f, if we are
able to get g such that either this inequality
holds for all x or this inequality holds for
all x, then the behavior of the integral can
be judged and it will depend on the inequality.
If this inequality holds then if this integral
is convergent, then this integral will converge.
Similarly this inequality hold and this integral
will diverge then this integral will also
diverge for it. So, that is what we get it
from here.
Another test which we have or let us take
an example here, that is better that if we
take say for example, yes suppose, I take
example, , suppose I take for p is less than
or equal to 1, test the convergence 
of the integral , .
Now, we know for, , ok, but =. This is what?
So, it will go to infinity as v tends to infinity
because this is positive; is positive; . So,
this is infinity. So, it means this integral
will diverge. And 1 to e, it is a finite sum;
so obviously, 1 to e does not contribute;
if this diverges then the total thing will
diverge.
So, the right hand sides, so since the integral
diverges, therefore, by comparison test, integral
. So, this way, we can find out. Then another
one example is also, suppose I take this one,
say another example if we take, say ok. let
us take , where , so that this integral diverges.
So, again we see here solution. When, then
. Therefore, why? Because the , log of x to
the power p is lying between less than 1.
So, it will be positive quantity; log1= 0
of course. So, it is positive and then for
p, , . So, we get the ..
Now, 1 upon this thing log x to the power
p, therefore 1 by x log x to the power p is
greater than equal to 1 by log x. But what
is the 1 by log x, but integral e to infinity
1 by log x and this integral diverges; is
it not. So, this diverges, d x; it diverges;
so this integral will diverge. Therefore,
it diverges; therefore, this diverges; e to
infinity to the power p d x diverges. So,
that is what the comparison test will be very
useful, but only thing is we have to identify
the relation. If we are unable to identify
such a relation, either fx is less than equal
to gx or fx is greater than equal to gx throughout
the interval, then only we can decide about
the nature, otherwise not.
The second test is mu test. What basis is,
if x to the power mu fx is bounded for x greater
than a, then integral a to infinity fx dx
is absolutely convergent, provided mu is greater
than 1 and diverges, and if let us, or we
can say in particular, if the limit of this,
limit of x to the power mu fx when x tends
to infinity exists, when mu exists, where
mu is greater than one.
Then the integral a to infinity fx dx converges
absolutely. Similarly, if limit of this 
x to the power mu fx when x tends to infinity
exists and is not equal to 0, different from
0, where mu is less than or equal to 1, then
the integral a to infinity fx dx diverges.
Of course, proof is simple.
What we do is, consider this modulus integral
a to x fx dx. Now this is less than equal
to a to x mod of fx dx. Now, what is given
is that if x to the power mu fx is bounded,
so modulus of this is less than equal to say
k bounded; that is, if this is less than equal
to k, that is x to the power mu or mod x to
the power mu fx is less than equal to k; k
is some constant, for some constant k.
So, we can write mod of fx is less than equal
to k times integral a to x, x to the power
minus mu dx. Now this integral is of the first
type; integral a to infinity x to the power
n type, minus n type. So, if n is we have
seen that this integral will converge when
mu is greater than 1. So, it converges; therefore,
the integral converges when mu is greater
than 1; is it not? When mu is greater than
1, when x tends to infinity this that is this
integral a to infinity fx dx, this integral
will converge. Here, this integral converges;
therefore, this integral will converge. So,
we get that thing. Similarly, if we take the
limit this in particularly limit of this exist
and if greater than 1, then this integral
converges absolutely. So, opposite of this
will be the diverging one.
The third test, of course it is not very useful,
but I will just simply say, if fx if integral
a to infinity fx dx, this converges absolutely
and gx is bounded and integerable in the interval
a to x, then for each x then integral a to
infinity fx gx dx is absolutely absolutely
convergent. Why? Because the reason is, consider
modulus integral xi xi dash fx gx dx necessary
and sufficient consider for the convergence
of the integral. So, this is less than equal
to k times integral xi to xi dash mod of fx
dx because this is given to be bounded; so
let it be bounded by k and this. Now, if this
integral fx converges absolutely is given;
so this is fine; less than equal to some quantity,
ok. So, this will be less than epsilon. So,
this shows necessary less than epsilon and
this shows that this is convergent epsilon.
Now, these are the general case. In particular,
we say the convergence of the also case.
So, in particular, we can see the cases when
convergent test for improper integral. This
is a particular case, ok. So, suppose it is
sufficient to consider. Let fx suppose let
fx is unbounded, unbounded at x equal to a;
improper integral of kind two. Of course,
x is in the interval a b; then this integral
a to b fx dx we can write as limit epsilon
tends to 0 a plus epsilon to b fx dx and this
if this limit exist, then we say improper
integral of kind two exist or convergent.
So, instead of going for the limit, we can
go by the test and the first test is comparison
test.
What this test says? If fx and gx are positive
are positive and fx is always less than equal
to gx in the interval a plus epsilon to b
in this interval, then the integral a to b
fx dx is convergent is convergent when integral
a to b gx dx is convergent. And the proof
is just by the previous thing because if you
take this thing a to b fx dx, then this is
less than equal to gx and gx less than equal
to a plus epsilon when epsilon tends to 0,
this will converge; therefore, this integral
will converge. So, it is not that.
The second test is the mu test. The mu test,
if x minus a to the power mu, fx is bounded
is bounded in the interval ab, then integral
a to b fx dx is absolutely convergent, if
provided mu is is strictly less than 1 and
diverges and diverges if mu is greater than
or equal to 1. So, this will be our …
Or equivalently we say in terms of the limit
or equivalently or equivalently in particular
if the limit exists, then if limit of this,
as x tends to limit of this as x tends to
a plus 0 x minus a to the power mu fx exists,
where mu is less than 1, then integral a to
b fx dx converges absolutely, ok. And in the
same way, in the same way, if limit of this
x minus a to the power mu fx when x tends
to a plus 0, fx exist and is and is not zero
and is not zero, where mu is greater than
or equal to 1, then the integral a to b fx
dx diverges. So, these tests are very important
and we will make use of these tests very frequently.
Let us see some problems, discuss the convergence,
the convergence and divergence of the integral
of the integral 0 to 1 x to the power n minus
1 log of x dx. Now, this integral if you see,
at x equal to 0, the integral because n, if
it is less than 1, then this will come in
the denominator; so at x equal to 0, it is
not defined; unbounded. Therefore, we can
discuss is into two phases: when n is lying
between 0, less than 0, when x is, n is equal
to 1, and when n is greater than 1. These
three cases will be there.
So, let us see the solution. Case one: When
n is equal to 1: Just first, simple case first.
What happened when n is equal to 1? This integral
0 to 1 reduces to 0 to 1 log x dx. Now, this
we can write it as epsilon limit epsilon tends
to 0 because when we take x tends to 0, it
is going to be unbounded. So, let epsilon
tends to 0, epsilon to 1 log of x dx integrate
by part. So, when you integrate by part, we
are getting the value of the integrate by
parts.
So, we get from here is limit epsilon tends
to 0; first function integral of the second;
first function integral of the second say
first function is log x; integral of the second
is x minus integral dc of the first; first
is 1 by x integral of the second this, and
then epsilon to 1 and like this. So, when
you are again integrating, finally you are
getting limit epsilon tends to 0; you will
get the value to be epsilon minus epsilon
log epsilon minus 1; is it not? That means
epsilon 1, 1 minus and then we are getting
to be this minus integral; this will be the
first integral. So, here is x minus 1 and
then plus epsilon and here epsilon log epsilon.
Then first integral of the second; so you
are getting this way; is it not? Because log
1 is 0 this is 0 to 1, I am sorry this is
0 to sorry epsilon to be 1. So, we get this
now when epsilon tends to 0, this is 0.
What about this? This we know that x to the
power alpha log of x when x tends to 0 is
always be 0. Why? Where alpha is greater than
0 when alpha is greater than 0 because by
using the L’hospital’s rules; if we apply
the L’hospital’s rule, you will get the
immediately the things because this can be
written as because this is 0 into infinity
form; 0 into infinity form; so it can be written
as logx divided by x to the power minus alpha
and limit x tends to 0. Now, when you differentiate
apply the L’hospital’s rule. So, logx
is 1 by x; this will give minus alpha x to
the power alpha minus minus 1. So, when you
get from here is and limit x tends to 0. So,
limit x tends to 0. This is minus alpha and
then x to the power alpha will go up; otherwise
a; so it is basically 0. So, this is always
be 0; it means this is 0. So, the limit of
this comes out to be minus 1.
It means the value of this integral is comes
out to be finite. Therefore, this converges.
So, integral 0 to 1 log x dx convergent is
convergent.
Now, case two: When n is greater than 1. Case
two: When n is greater than 1, n is say lying
between 0 and 1. If integral proper… So,
when n is greater than 1, the integral integral
0 to 1, x n minus 1 log x dx. Now, here, the
integrand when n is greater than 1, this basically
comes in the numerator only. This comes to
in numerator; therefore, the integral is proper.
This integral is proper integral because the
limit of this x n minus 1 log x as x tends
to 0 is 0. So, this limit is 0; means, it
is finite; does not go to unbounded. So, it
is a case of proper integral. So, it has be
convergent; so convergent.
So, nothing to prove when n is less than 1,
n is less than 1. Then let us denote let fx
stands for x to the power n minus 1 into log
x. Now, consider x to the power mu fx. Now,
one thing is when you take x to the function
fx, what is the result? If you remember the
result was mu test. The mu test was, yes if
this is bounded in the interval, bounded in
a b, then this is absolutely convergent provided
mu is less than 1. So, we can use this criteria
here.
So, when n is less than this, in this x to
the power mu dx fx converges; converges when
mu plus n minus 1 is greater than 0 because
this is equivalent to what? This is equivalent
to x to the power mu plus n minus 1 into fx;
fx means log x; log x that is and we know
the limit of this is 0 if this is greater
than the positive quantity. So, the limit
of this, it will converge. Therefore, this
converges when mu plus n minus 1 is greater
than 0. So, this implies that is the 1 minus
n is less than mu, but integral converges,
but this integral 0 to 1 x to the power mu
fx dx.
So, this is a bounded function. So, this integral
converges by mu test because it is equal to
the same as integral 0 to 1 x minus 0 mu fx
dx. So, this integral converges if mu is strictly
less than 1. So, from here, this show 1 minus
n is strictly less than 1; this implies n
is greater than zero. So, this integral converges
when n is greater than equal to 0, greater
than 0. Already we have seen when n is greater,
lying between less than equal to 1 is convergent
and diverges when n is less than equal to
0.
So, what we see here, therefore, integral
0 to 1 x to the power n minus 1 log x dx convergent
for 0 less than n less than equal to 1 and
divergent for n less than or equal to 0 because
when mu is greater than equal to 1 it is diverging.
When mu is greater than equal to 1 means n
is less than equal to 1; so divergent. So,
that is the result.
Another example, we will discuss discuss the
convergence discuss the convergence of the
following integrals following integrals. The
first integral is 0 to infinity x to the power
alpha minus 1, e to the power minus x dx and
alpha is greater than 0. In fact, this is
denoted by gamma alpha. So, we will use the
later on, but gamma alpha. Gamma alpha second
integral say is 0 to 1 x to the power m minus
1 1 minus x to the power n minus 1 d x when
m is greater than 0 n is greater than 0. So,
this integral we denoted by the gamma beta
m n; m n is the beta function; beta m n - this
converges here. So, this basically when alpha
greater than 0, this will converge. When m
is greater than 0, n is greater than zero;
so basically this is convergent; this is convergent;
this we will test it.
Solution is, first one: Now, we look this
integral; alpha is greater than 0; so when
alpha is lying between 0 and 1, this x will
come in the denominator. So, it becomes unbounded
when x is 0 at the lower limit. And then because
of the upper limit, it is mixed; again a improper
integral. So, we will divide this integral
1. 0 to infinity x to the power alpha minus
1 e to the power minus x dx as 0 to 1 x to
the power alpha minus 1 e to the power minus
x dx plus 1 to infinity x to the power alpha
minus 1 e to the power minus x dx. So, this
is first integral; this is second integral;
two parts. Now, check the first part first
integral.
Let us denote let x to the power alpha minus
1 e to the power minus x is fx. So, what we
see here, we see that x to the power one minus
alpha fx; that is equal to what? e to the
power minus x and that goes to one as x tends
to 0 therefore, this integral is convergent;
therefore, integral this fx dx will converge.
therefore, this integral fx dx x to the power
alpha minus 1; this integral converges; alpha
minus 1 e to the power minus x 0 to 1 the
d x converges when 1 minus alpha is less than
1; that is alpha is greater than 0.
The second part of this, second part of this,
the function limit of this x to the power
alpha minus 1 e to the power minus x when
x tends to infinity, this is always be 0.
Whatever the alpha may be, this power x to
the power e to the power x will, minus x,
will go in the denominator; so it will be
always 0; therefore, whatever the value of
whatever be the value of alpha; therefore,
the second integral 1 to infinity x to the
power alpha minus 1 e to the power minus x
dx will be convergent when alpha is positive.
So, this will be greater than 0; therefore,
this integral converges for all.
Now, second part is full; second integral
is again 0 to 1 x to the power m minus 1 1
minus x raised to the power n minus 1 dx.
Here, again it depends on m and n suppose
m is less than 1, then at this point, it is
divergent and when n is less than 1 at the
upper point it is again divergent. So, what
we do? We break up into two parts 0 to alpha
x to the power m minus 1 1 minus x 
n minus 1d x and plus alpha to 1 x to the
power m minus 1 1 minus x n minus 1 dx. So,
we will look the convergence of both; both
integral converges then only this integral
will converge.
So, let us see the first where alpha is where
alpha is, obviously, is lying between 0 and
1. now let us see the first part. denote x
to the power m minus 1 1 minus x to the power
n minus 1 is suppose fx. Then what we see
here is, we see that the limit of this function
fx when multiplied by 1 minus m fx as x tends
to 0, is basically what? This is the same
as limit x tends to 0 1 minus x to the power
n minus 1; so that comes out to be 1; it means
it is finite. So, the first integral therefore,
the integral 0 to alpha x to the power m minus
1 1 minus x n minus 1 dx will converge provided
what? Provided this 1 minus m, by which you
are multiplying, is strictly less than 1;
that is m is greater than zero by mu test.
So, this converges, provided this.
Now, since again second part is since limit
of this 1 minus x to the power 1 minus n fx
as n x tends to 1 is nothing but what? This
will be the same as 1 minus x to the power
n minus m minus 1 as x tends to 1. So, it
comes out to be 0. Why 1 minus x tends to
1 fx what is our fx? fx is denoted by this
one; is it not? So, when you multiply this
by 1 minus x 1 minus n. So, that becomes 0
and when x to the sorry this will go n minus
1 plus 1 minus n and then multiplied by x
to the power 1 minus x to the power m minus
1. So, this limit comes out to be 1.
So, once this limit comes it means it is a
finite quantity. Therefore, the second part
will be convergent. So, second part alpha
to 1 x m minus 1 1 minus x n minus 1 dx converges
if this 1 minus n is strictly less than 1;
that is n is greater than 0. Therefore, whole
integral 0 to 1 x m minus 1 will converge
for m is greater than 0, n is greater than
0 because this is true for all n and for all
m; n is greater than 0 for all m and earlier
m is greater than 0 for for all n. So, common
factor is m is greater than 0 and n is greater
than 0 and this is known as the beta function.
So, it will also be very useful function.
Apart from this, we will now look few more
examples that is also directly following...
Let us see for p greater than 1, for p greater
than 1, is integral 1 to infinity log x over
x to the power p dx convergent? This is our
goal. Now, this we will show it by first definition
itself. So, let us suppose we can write it
this for convergence part is 1 to infinity
log x over x to the power p dx can be written
as limit v tends to infinity 1 to v log x
over x to the power p because x varies from
1 to infinity; so it is well defined at this
point; 0 is not available. Only point when
x tends to infinity, this because of the infinity
it is improper. So, if this limit exists then
we say the integral will be convergent. Now,
this you integrate by part by part by parts.
So, if you apply the integration by parts
taking the first function, this second function
this.
So, we get first function first function is
1 upon x to the power p; first function is
sorry log x. First function integral of the
second. So, integral x to the power minus
p plus 1 over 1 minus… So, we get 1 minus
p x to the power minus p plus 1. So, p minus
1 first function integral of the second; integral
of the second is log of minus p; that is 1
by p first function integral of this; integral
dc of the function dc of the first 1 by x
and integral of the second will be 1 minus
p x to the power p minus 1 x to the power
p minus 1 and that will be equal to what?
dx. So, when we again apply this integration,
now it is simple integration.
So, when we get this integration, what we
get from here is that this comes out to be
limit v tends to infinity 1 over 1 minus p
log of v over log of v over v to the power
p minus 1 minus minus 1 over 1 minus p whole
square into 1 by v p minus 1; then minus minus
1 over 1 minus p square. just simple integration
you will see. So, what we get when v tends
to infinity? Now, this part, only thing, limit
of this is because this will go to 0. So,
what we claim is the limit of this will be
0; limit of this will be 0. So, basically,
the value will come out to be 1 over 1 minus
p whole square as limit of this limit of log
v over v to the power p minus 1 v tends to
infinity is 0 by L’hospital’s rule. That
you can just verify as value p greater than
1; so this proves that. So, the integral converges,
integral converges.
Now, next, if we look this again, integral
second, evaluate this integral integral 0
to infinity e to the power minus x cosine
x dx. So, what we do is we are first taking
this as the limit v tends to infinity integral
0 to v e to the power minus x cos x dx. Now,
what I say, we know that if I take I to be
the integral e to the power minus x cos x
dx and then taking first and second function,
first function integral of the second minus
integral dc to first integral integral of
the second minus Integral of the second is
sin x minus integral dc of the first and integral
of the second is sin x dx and again first
function and second function, and again repeat
this thing; what we get? We will get the I
again. So, finally, we are getting I. So we
do it again. So, what we get? I comes out
to be half half e to the power minus x sin
x minus cosine x. So, now 0 to infinity of
this means I 0 to v is; is it not? So, this
will be…. Therefore, what we want is the
limits.
So, we get from here is hence 0 to infinity
0 to infinity e to the power minus x cos x
d x that will be equal to what limit? limit
v tends to infinity is half e to the power
minus v minus v cosine of v minus sin of v
minus minus 1. because this will be when you
go for this 0 to v 0 to v. And then finally,
you will get something here. Integration sorry
integration when you do it, this part when
you do for integration you are getting this
thing minus v and 0.
So, finally, integration will be from here;
this will be 0 to v. When you are writing
integral 0 to v e to the power minus x cos
x d x, then you will get this result, which
while I am doing this one; half of this this
is half; e to the power minus 
v e to the power minus v cosine v minus sin
v and then then minus minus plus half. This
you will get it. So, when you take the limit
of this v tends to 0 to infinity get this
thing. So, we get from here, what we get is
now, the answer will be 
when 
v tends to 
infinity this is bounded function; bounded
by 2. mod cos x is less than cos v is less
than equal to 1; mod of this less than v equal
to 1. So, this part will tend to 0 and here
you are getting is half; so that is the answer;
so convergent; convergent one; so this we
get it. Now, one more example is, let us evaluate
integral 0 to infinity e to the power minus
root x dx. So, here, if we substitute root
x equal to t square, then everything will
be fine and we get 0 to infinity e to the
power minus t and twice t dt into 2 t dt 2
t dt and that is all. So, when you integrate
you are getting the value 2. So, it is convergent
and that is it.
Thank you very much.
Thanks.
So, this completes your course. Fine.
