I tried it this way........
I = ∫ (X/1! - x^3/3! + x^5/5! .....)/x dx
= ∫ (1- x^2/3! + x^4/5! - .....) dx
= (x - x^3/3.3! +x^5/5.5!....)
Applying limits now is troublesome????????
I tried it this way........
I = ∫ (X/1! - x^3/3! + x^5/5! .....)/x dx
= ∫ (1- x^2/3! + x^4/5! - .....) dx
= (x - x^3/3.3! +x^5/5.5!....)
Applying limits now is troublesome????????
u know wat i guess this is a non integrable fuction ....with limits from -∞ to + ∞ means an indefinite integration.....and integral sinx/x is a non integralbe function
Assume : I(a)=2\int_{0}^{\infty}{\frac{e^{-at}sint}{t}}dt
\frac{dI}{da}=-2\int_{0}^{\infty}e^{-at}sint\: dt
\frac{dI}{da}=2\left[\frac{e^{-at}(asint+cost)}{a^{2}+1} \right]_{0}^{\infty}
\frac{dI}{da}=-\frac{2}{a^{2}+1}
I(a)=-2\, tan^{-1}a+c
as a tends to infinity integral tends to 0 : 0=-2\, tan^{-1}\infty + c
c=\pi
I(a)=-2\, tan^{-1}a+\pi
required integral : I(0)=\pi
uttara!!
he is much elder thn U.. see his profile.. so calling him dubey.. :-o