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### Integrating sin(x)/x

```Date: 02/16/2005 at 12:30:05
From: Shawn
Subject: Limit of a special integral

I'm trying to find the integral from 1 to t of sin(x)/x so that I can
calculate the limit as t goes to infinity.

I either get to:

sinx*lnx - integral(cosx*lnx) or -cos(x)/x - integral(cos(x)/x^2)

Neither of those results do I see as coming back to anything I can use
to find a final integral, let alone find useful for calculating the
limit.

```

```
Date: 02/18/2005 at 00:19:05
From: Doctor Ricky
Subject: Re: Limit of a special integral

Hey Shawn,

Thanks for writing Dr. Math!

This integral is special since you can't use integration by parts to
simplify it to a function of x not involving an iterated integral.
However, we can find the limit of this function two separate ways.

First, we can find the limit as t->00 (infinity) of sin(t)/t by
inspection of the two separate functions, g(t)=sin(t) and h(t)=t.

We know g(t) is a periodic function, since its period is constant and
it follows the exact same curve scheme if you isolate two distinct
periods of the curve (i.e. sin(t) will curve the same between [pi,
2*pi] as it does between [3*pi, 4*pi]).

Since it is a periodic function, the limit of g(t) as t->00 never
approaches any particular value since it takes on different g(t)
values for each t value as t increases.  Therefore the limit of
g(t)/h(t) is undefined, since the limit of the division of two
functions is the same as the division of the limits, i.e.

lim   [  g(t)  ]      lim    g(t)
t->00  [ ------ ] =   t->00           D.N.E
[  h(t)  ]     ------------ =  ----- = Does Not Exist
lim    h(t)      00
t->00

We can also show this using L'Hopital's Rule, which states that if a
limit of a function divided by another function is in indeterminant
form (0/0, 00/00, 00/0 or D.N.E), then we can take the derivative of
the numerator and divide it by the derivative of the denominator (NOT
the Quotient Rule!) and this will equal our limit of our original
equation.

This means:

lim   [  g(t)  ]              lim    g'(t)
Since     t->00  [ ------ ] = DNE, find t->00  ------
[  h(t)  ]                     h'(t)

Since g'(t) = cos(t) and h'(t) = 1, now our limit is:

lim   [  g'(t)  ]    lim    cos(t)
t->00  [ ------  ] = t->00   ------ = D.N.E since cos is periodic
[  h'(t)  ]              1

Since any successive derivative of g'(t) will give us another periodic
function and the derivative of h'(t) is 0, we will never have an
answer that exists and therefore our limit will never approach any
value as t approaches infinity (in other words, as t increases without
bound).  Therefore, our limit does not exist.

Hope this helps, but if you have any more questions or concerns, feel
free to write back!

- Doctor Ricky, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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