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