x - \(\frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.6}+....\)

 = x (1 - \(\frac{x^2}2 + \frac{x^4}{2^2(1.2)}-\frac{x^6}{2^3(1.2.3)}+....\) )

 = x(1 - \(\frac{x^2}2+\cfrac{(\frac{x^2}2)^2}{2!}-\cfrac{(\frac{x^2}2)^3}{3!}+....\))

 = x e^{\(-\frac{x^2}2\)} (\(\because\) e^-x = 1 - x + \(\frac{x^2}{2!} - \frac{x^3}{3!}+...\))

\(\therefore\) \(\int\limits_0^{\infty}(x - \frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.7}+...)\) e^{\(\frac{x^2}4\)} dx

 = \(\int\limits_0^{\infty}xe^{-\frac{x^2}2}.e^{\frac{x^2}4}dx\) 

 = \(\int\limits_0^{\infty}xe^{\frac{x^2}4-\frac{x^2}2}\)dx

 = \(\int\limits_0^{\infty}xe^{-\frac{x^2}4}\)dx

Let \(\frac{x^2}4=t\)

Then \(\frac{2xdx}4dt\)

⇒ xdx = 2dt

limit converts into from t = \(\frac{0^2}4 = 0\) t = \(\frac{\infty^2}4=\infty\)

 = \(2\int\limits_0^{\infty}e^{-t}dt\) 

 = \(-2[e^{-t}]_0^{\infty}\)

 = -2(e^-\(\infty\) - e⁰)

 = -2 (\(\frac{1}{e^{\infty}}-1\))

 = -2 (\(\frac1{\infty}\) - 1)

 = -2(0 - 1)

 = -2(0 - 1)

 = -2 (-1)

 = 2

Hence, \(\int\limits_0^{\infty}(x - \frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.7}+...)\)e^{\(\frac{x^2}4\)} dx = 2