# Finding $\int^{\infty}_{0}\bigg(\frac{1-\cos 7x}{x}\bigg)e^{-x}dx$

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

• It should be $$\frac{\log (50)}{2}$$ – Dr. Sonnhard Graubner Jun 13 at 13:54
• Expanding in a power series is a good idea, but you should only expand the left term in the integrand, not the exponential. You also messed up the first term after expanding it. – Cameron Williams Jun 13 at 13:56
• math.stackexchange.com/questions/1807410/… – Olivier Oloa Jun 13 at 14:11
• @Dr. Sonnhard Graubner , you are right, you may see Answer no. 3. – Dr Zafar Ahmed DSc Jun 13 at 17:28

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 $$I(a)=\frac{1}{2} \log \left(\frac{49}{a^2}+1\right).$$
Use $$1/x=\int_{0}^{\infty} e^{-tx} dt$$, then $$I=\int_{0}^{\infty}\int_{0}^{\infty} (1-\cos 7x) e^{-x(1+t)} dt~ dx=\Re \left(\int_{0}^{\infty}\int_{0}^{\infty}[e^{-x(1+t)}-e^{-x(1+7i+t)}] dx ~ dt \right)=\Re \left(\int_{0}^{\infty} \left( \frac{dt}{1+t}- \frac{dt}{1+7i+t)} \right) \right).= \Re \left( \ln\frac{1+t}{1+7i+t}\right)_{0}^{\infty}= -\Re \left(\ln \left[\frac{1}{1+7i}\right]\right)=\frac{\ln 50}{2}.$$