What is #int_1^oo sinx/x^2 dx#?

Question

What is #int_1^oo sinx/x^2   dx#?

Isaiah Allen · Accepted Answer

#int_1^oo sin(x)/x^2\ dx=sin(1)-Ci(1)~~0.504# To work out the antiderivative, we'll apply integration by parts with #f=sin(x)# and #g'=x^-2# This makes our integral: #-sin(x)/x-int-cos(x)/x\ dx=-sin(x)/x+int\ cos(x)/x\ dx# On the right there we have a special integral, the Cosine Integral. It is usually denoted using #Ci(x)#. This means the answer to the integral is: #Ci(x)-sin(x)/x+C# Now we can plug in our limits of integration to get: #int_1^oosin(x)/x^2=lim_(x->oo)(Ci(x)-sin(x)/x)-(Ci(1)-sin(1))# Because the #Ci(x)# function is the result of an integration, it will have some constant. What this constant is doesn't really matter, because in most applications you will have it cancel since you subtract two #Ci(x)# values (due to the Fundamental Theorem of Calculus). When computing the #Ci(x)# function, you have to pick some value for this constant (you can choose whatever you like for convenience, since it will cancel). You might expect the choice to be #0#, but it has been decided that this constant should be #gamma# (the Euler Mascheroni constant). This is so that: #lim_(x->oo)Ci(x)=0# This means we can evaluate the limit in our expression to be (I expect that you are able to evaluate the sine limit on your own): #lim_(x->oo)(Ci(x)-sin(x)/x)=0# So, we're left with: #-(Ci(1)-sin(1))=sin(1)-Ci(1)# You could just stick the #Ci(1)# value in something like Wolfram Alpha to get an answer, but just for fun, let us look at how one might arrive at the value. The simplest way to approximate #Ci(1)# is probably using a Maclaurin series. We know that: #Ci(x)=int\ cos(x)/x\ dx# The classic technique here is to work out a series for #cos(x)/x# and then integrate it. We know the Maclaurin series for cosine, so we can just divide all of the terms by #x#: #cos(x)/x=1/xsum_(n=0)^oo(-1)^nx^(2n)/((2n)!)=sum_(n=0)^oo(-1)^nx^(2n-1)/((2n)!)=# #=1/x-x/(2!)+x^3/(4!)-x^5/(6!)...# To now integrate this, we can treat the first value specially (since it will have an integral of #ln(x)#), and we can use the reverse power rule for the rest: #int\ sum_(n=0)^oo(-1)^nx^(2n-1)/((2n)!)\ dx=C+ln(x)+sum_(n=1)^oo(-1)^nx^(2n)/(2n(2n)!)# #C# is the constant of integration, and we previously picked this to be #gamma#, so we must do the same here (otherwise they wouldn't cancel): #Ci(x)=gamma+ln(x)+sum_(n=1)^oo(-1)^nx^(2n)/(2n(2n)!)# Now we can use this formula to evaluate #Ci(1)#. I wrote a small Java program to evaluate the series up to #15# terms, and getting a value of: #Ci(1)~~0.3374039# Now we just subtract this from #sin(1)# to get roughly: #sin(1)-Ci(1)~~0.504#

Elijah Abram · Answer

undefined The integral $ \int_1^\infty \frac{\sin(x)}{x^2} \, dx $ converges and its value is approximately $ 0.45 $.