Interesting definite and improper integralsINTRODUCTIONMost of the integrals that are analytical have proved to have interesting applications in the outside world. Yet, there are integrals which have, important applications in the real world, but are counter-intuitive. These integrals often do not have an analytical solution in terms of standard functions, but have a finite value for the improper integral. For instance the integrals, ∫0∞(lnx)2(x2+1)2dxand∫0∞lnx(x2+1)dx, are not analytical in terms of standard functions, yet, astonishingly, a finite value is obtained for the improper integrals ranging from 0 to infinity. In contrast, an analytical solution to theintegral∫tannx dx, where n is a real number, could be found by employing special methods. However, it is not at all obvious how such an integral could produce an analytical solution. These observations left me baffled and it came to thought that such integrals would prove interesting to investigate. In this exploration I would, in the first section, discuss the integral of∫tannx dx, specifically whenn is a rational number, and attempt to find a value for the improper integral from 0 to infinity. In the second section I would prove that an analytical solution for ex2and cosxxcannot be arrived at, in terms of standard functions, but I will attempt to provide an alternate solution using series
expansions. Consequently, I would explore different methods to find the value for the improper integrals. SECTION1Problem 1let’s consider the functionf(x)=tannx, where n∈ℝ.For n=1, the standard tangent function could be integrated to give –ln|cosx|.This could be simplified to giveln|secx|. However, when n approaches rational numbers; the integral turns rather complex. Let us then let n be12. The integral could be expressed as ∫√tanx dx.The substitution for suchan integral is very hard to deduce. In order to evaluate this integral we will define two supplementary functions, I1and I2, such that the sum or the difference would result in the integral we are interested in.
Let:I1be √tanx(¿+√cotx)dx∫¿I2be(√tanx−
