# Evaluate $\int_1^\infty \frac{\cos x}{x}dx$.

I have a doubt, I know that this integral doesn't have an antiderivative but, is there a way to solve this:

$$\int_1^\infty \frac{\cos x}{x}dx$$

Using Laplace transform or something similar? I say Laplace because I know how to solve this by Laplace transform: $$\int_1^\infty \frac{\sin x}{x}dx$$

• What have you tried? Have you tried integration by parts? Show some of your work so other users can help appropriately. – Test123 Oct 14 '17 at 19:04
• It's the cosine integral. – Simply Beautiful Art Oct 14 '17 at 19:05
• see also here mathworld.wolfram.com/SineIntegral.html – Dr. Sonnhard Graubner Oct 14 '17 at 19:12
• Laplace transform is from 0.$\int_0^\infty \frac{\sin x}{x}dx$ – Nosrati Oct 14 '17 at 19:15
• What does it mean to solve an integral? $$\int_{1}^{+\infty}\frac{\cos x}{x}\,dx = -\text{Ci}(1) = \int_{0}^{+\infty}\frac{s\cos(1)-\sin(1)}{1+s^2}\,e^{-s}\,ds \approx -\frac{1}{3}.$$ – Jack D'Aurizio Oct 14 '17 at 19:34

$$\int_1^\infty \frac{\cos x}{x}dx=\operatorname{Ci}(1)=-\gamma+\int_0^1\frac{1-\cos t}{t}dt$$ and ‎\begin{eqnarray*}‎ ‎{\cal L}\Big(\int_0^x\frac{1-\cos t}{t}dt\Big) &=& \frac{{\cal L}\Big(\dfrac{1-\cos t}{t}\Big)}{s}‎ \\‎ ‎&=& \frac{\int_s^\infty\dfrac1s-\dfrac{s}{s^2+1} ds}{s}‎ ‎\\‎ ‎&=& \frac1s\Big[\ln\frac{s}{\sqrt{s^2+1}}\Big|_s^\infty\Big] \\‎ ‎&=& \frac{-1}{2s}\ln\frac{s^2}{s^2+1} \\‎ ‎&=& \frac{1}{2s}\ln\Big(1+\frac{1}{s^2}\Big)‎ ‎\end{eqnarray*}‎ so ‎\begin{eqnarray*}‎ \int_0^x\frac{1-\cos t}{t}dt &=& {\cal L}^{-1}\left(\frac{1}{2s}\ln\Big(1+\frac{1}{s^2}\Big)‎ \right) \\ &=& {\cal L}^{-1}\left(\frac{1}{2s}\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{n}\Big(\frac{1}{s^2}\Big)^n‎ \right) \\ &=& {\cal L}^{-1}\left(\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{2n}\dfrac{1}{s^{2n+1}}‎ \right) \\ &=& \sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{2n}\dfrac{x^{2n}}{(2n)!}‎ ‎\end{eqnarray*}‎ then $$\int_1^\infty \frac{\cos x}{x}dx=\color{blue}{-\gamma+ \sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{2n(2n)!}}$$