Give an example of a function which is not integrable on [0,π].Justify your answer.?

For a function to be non-integrable on a certain interval, we require that function to have a vertical asymptote on that interval. When thinking of the definite integral as the area under the curve, a vertical asymptote within the interval (or at one of the endpoints) stops us from taking the area right away and forces us to split our integral into separate intervals in the case of a discontinuity within the interval.

Think of a function that has a vertical asymptote on the interval #[0,pi]#.

Consider the function #tan(x)# on the interval #[0,pi]# .

#tan(x)# has vertical asymptotes at #x=(pin)/2# where #n# is any integer. This is because #tan(x)=sin(x)/cos(x)#, and #cos(x)=0# for #x=(pin)/2#, thus giving us a division by #0# error for #tan(x)#.

#int_0^pitan(x)dx# is not integrable due to the vertical asymptote at #x=pi/2#; #tan(x) # meets our requirements.

Any other function with a vertical asymptote on #[0,pi]# works as well, such as

#f(x)=1/(x-pi/2), f(x)=5/(2x-pi), f(x)=(-3x)/(4x-2pi)#.

Infinitely many functions could meet this requirement, but #tan(x)# is a common function which cannot be integrated over #[0,pi]#.