# 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 $$I=\int_{0}^{\infty}\int_{0}^{\infty} (1-\cos 7x) e^{-x(1+t)} dt~ dx=\Re \left(\int_{0}^{\infty}\int_{0}^{\infty}[e^{-x(1+t)}-e^{-x(1+7i+t)}] dx ~ dt \right)=\Re \left(\int_{0}^{\infty} \left( \frac{dt}{1+t}- \frac{dt}{1+7i+t)} \right) \right).= \Re \left( \ln\frac{1+t}{1+7i+t}\right)_{0}^{\infty}= -\Re \left(\ln \left[\frac{1}{1+7i}\right]\right)=\frac{\ln 50}{2}.$$
$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\bracks{1 - \cos\pars{7x} \over x}\expo{-x}\dd x} = \Re\int_{0}^{\infty}\bracks{1 - \expo{7\ic x} \over x}\expo{-x}\dd x \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,& -\Re\int_{0}^{\infty}\ln\pars{x}\bracks{-\expo{-x} + \pars{1 - 7\ic}\expo{-\pars{1 - 7\ic}x}}\dd x \\[5mm] = &\ \int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x - \Re\int_{0}^{\pars{1 - 7\ic}\infty}\bracks{\ln\pars{x} - \ln\pars{1 - 7\ic}}\expo{-x}\dd x \\[5mm] = &\ \Re\ln\pars{1 - 7\ic} = \bbx{{1 \over 2}\,\ln\pars{50}} \approx 1.9560 \end{align}