Finding $\displaystyle \int^{\infty}_{0}\bigg(\frac{1-\cos 7x}{x}\bigg)e^{-x}dx$
Plan
$$I =\int^{\infty}_{0}\frac{1}{x}\bigg(1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}+\cdots \bigg)\bigg(1-\frac{x}{1!}+\frac{x^2}{2!}+\cdots \bigg)dx$$
How do i solve it Help me please
Best Answer
Hint. Assume $a>0$. Set $$ I(a):= \int^{\infty}_{0}\bigg(\frac{1-\cos 7x}{x}\bigg)e^{-ax}dx $$ then, by differentiating with respect to $a$, one gets $$ I'(a)= -\int^{\infty}_{0}\bigg(1-\cos 7x\bigg)e^{-ax}dx=-\frac{1}{a}+\frac{a}{a^2+49} $$ giving $$ I(a)=\frac{1}{2} \log \left(\frac{49}{a^2}+1\right). $$
Hope it helps.