Till now, we have learned integrals 
dx#card=math&code=%5Cint_a%5Eb%20f%28x%29dx&id=a5JBZ) where 
is finite, and 
is continuous over this closed interval.
Now we extend 
to an infinite interval (but 
is still continuous). This is
Infinite (interval) intergrals
By the Newton-Liebniz formula 
dx%20%3D%20F(t)%20-%20F(a)#card=math&code=%5Cinta%5Et%20f%28x%29dx%20%3D%20F%28t%29%20-%20F%28a%29&id=MVGQY), since 
is continuous 
#card=math&code=%5Ba%2C%20%2B%5Cinfty%29&id=iqz8w) , we deduce that 
#card=math&code=F%28x%29&id=bdRzo) is also continuous over #card=math&code=%5Ba%2C%20%2B%5Cinfty%29&id=TK4wD) . Then we can consider the limit %20-%20F(a)%5D%20%3D%20%5Clim%7Bt%5Cto%20%5Cinfty%7D%20F(t)%20-%20F(a)#card=math&code=%5Clim%7Bt%20%5Cto%20%5Cinfty%7D%20%5BF%28t%29%20-%20F%28a%29%5D%20%3D%20%5Clim_%7Bt%5Cto%20%5Cinfty%7D%20F%28t%29%20-%20F%28a%29&id=rFFpc).
Example of infinite integrals: Gabriel’s horn has finite volume but infinite surface area.
Now we look at the second kind of Improper integrals:
is continuous at 
(usually 
) except a point 
that (1)$ c =a $ or (2) 
or (3) 
lies in 
#card=math&code=%28a%2C%20b%29&id=AP6xS). And 
has an infinite discontinuity at 
.
Example (problem 305 on page 593):
%20%5Cfrac%7Bx%7D%7B1-x%5E2%7D%20dx#card=math&code=%5Cint_0%5E2%20%5Cfrac%7Bx%7D%7B1-x%5E2%7D%20dx%20%3D%20%20%5Cbig%28%5Cint_0%5E1%20%2B%20%5Cint_1%5E2%5Cbig%29%20%5Cfrac%7Bx%7D%7B1-x%5E2%7D%20dx&id=o0Qh6), we denote 
%3Dx%2F(1-x%5E2)#card=math&code=f%28x%29%3Dx%2F%281-x%5E2%29&id=wdiwb) and let 
.
Now as 
goes from 
to 
, 
will go from 
to 
. The range of 
is 
. But 
is not continuous at 
. Thm 5.8 fails to apply.
Conclusion: the original integral diverges (in other words, not existed).
Let us verify this statement via explicit computation. The antiderivative is
%20%5Cln%20%7C1-x%5E2%7C%2BC#card=math&code=%5Cint%20%5Cfrac%7Bx%7D%7B1-x%5E2%7D%20dx%20%3D%20%28-1%2F2%29%20%5Cln%20%7C1-x%5E2%7C%2BC&id=demrA). Using the Newton-Leibniz formula we see that
%20%3D%20-%5Cinfty#card=math&code=%5B%5Cln%20%7C1-x%5E2%7C%5D0%5E1%20%3D%5Clim%7Bx%20%5Cto%201-%7D%20%5Cln%281-x%5E2%29%20%3D%20-%5Cinfty&id=hRHzq),
, so
%20%5B%20%5Clim%7Bx%20%5Cto%201-%7D%20%5Cln(1-x%5E2)%20%2B%20%5Cln(3)%20-%20%5Clim%7Bx%20%5Cto%201%2B%7D%20%5Cln%20(x%5E2-1)%20%5D%20%3D(-1%2F2)%20%5Cln(3)#card=math&code=%28-1%2F2%29%20%5B%20%5Clim%7Bx%20%5Cto%201-%7D%20%5Cln%281-x%5E2%29%20%2B%20%5Cln%283%29%20-%20%5Clim%7Bx%20%5Cto%201%2B%7D%20%5Cln%20%28x%5E2-1%29%20%5D%20%3D%28-1%2F2%29%20%5Cln%283%29&id=wFXEU) (formally)
This value is the “principal value” for improper integral.
Another example of principal value (in the case of infinite integral):
Conclusion: this infinite integral does not exist. But we can talk about its principal value, which is zero via the “net signed area” interpretation.
Remark on problem 94 on page 545. 
%20%5Cfrac%7B%5Csin%20t%7D%7B1%2Bt%5E2%7D%20dt#card=math&code=%5Cbig%28%5Cint_%7B-%5Cpi%7D%5E0%20%2B%20%5Cint_0%5E%7B%5Cpi%7D%20%5Cbig%29%20%5Cfrac%7B%5Csin%20t%7D%7B1%2Bt%5E2%7D%20dt&id=WFCZa)
The integrand has antiderivatives, but cannot express explicitly in terms of elementary functions. To evaluate it, let 
. Then 
, and as 
goes from $-\pi $ to 
, 
will go from $\pi $ to 0. Thus
%20%7D%7B1%2Bt%5E2%7Ddt%20%3D%20%5Cint%7B%5Cpi%7D%5E0%20%5Cfrac%7B-%5Csin%20u%7D%7B1%2Bu%5E2%7D%20(-du)%20%3D%20%5Cint%7B%5Cpi%7D%5E0%20%5Cfrac%7B%5Csin%20u%7D%7B1%2Bu%5E2%7D%20du%20%3D-%5Cint0%5E%5Cpi%20%5Cfrac%7B%5Csin%20t%7D%7B1%2Bt%5E2%7D%20dt#card=math&code=%5Cdisplaystyle%20%5Cint%7B-%5Cpi%7D%5E0%20%5Cfrac%7B%5Csin%20%28t%29%20%7D%7B1%2Bt%5E2%7Ddt%20%3D%20%5Cint%7B%5Cpi%7D%5E0%20%5Cfrac%7B-%5Csin%20u%7D%7B1%2Bu%5E2%7D%20%28-du%29%20%3D%20%5Cint%7B%5Cpi%7D%5E0%20%5Cfrac%7B%5Csin%20u%7D%7B1%2Bu%5E2%7D%20du%20%3D-%5Cint_0%5E%5Cpi%20%5Cfrac%7B%5Csin%20t%7D%7B1%2Bt%5E2%7D%20dt&id=djr50).
Conclusion: 
.
