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Show that $$\int_0^x \frac{1-\cos(t)}{t}=\gamma+\ln(x)-\operatorname{Ci}(x)$$ where $$\operatorname{Ci}(x)=-\int_x^\infty \frac{\cos(t)}{t} \, dt$$ and gamma is an euler-mascheroni constant. I did as follows: $$\int_0^x \frac{1-\cos(t)}{t} \, dt=\int_0^x \frac{1}{t} \,dt-\int_0^x \frac{\cos(t)}{t}\,dt$$ $$=\int_0^x \frac{1}{t}\,dt-\int_0^\infty \frac{\cos(t)}{t}\,dt+\int_x^\infty \frac{\cos(t)}{t} \, dt=\int_0^x \frac{1}{t}-\int_0^\infty \frac{\cos(t)}{t}-\operatorname{Ci}(x)$$

I am stuck with the last two integrals. How do I proceed? A shorthand notation tells me that I get $\ln(x)-\ln(0)-\Gamma(0)$ which is wrong since $\ln(0)$ and $\Gamma(0)$ are undefined unless $$\ln(0)+\Gamma(0)=-\gamma$$

Adam S
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2 Answers2

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Here is an approach.

$$ \int_0^x \frac{1-\cos(t)}{t}dt = \sum_{k=1}^{\infty}\frac{(-1)^k}{(2 k)!}\int_0^x t^{2k-1}\,dt = \dots\,. $$

Note: The Taylor series of $\cos(t)$ is

$$ \cos(t) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}t^{2k}. $$

Mhenni Benghorbal
  • 47,431
  • Even though this is an interesting approach somehow this approach does not take me anywhere. – Adam S Aug 29 '14 at 01:25
  • Okay. I solved it using your approach. I expanded Ci(x) function that led me to Euler constant, natural log and series expansion. I used that series expansion to get that formula in question. – Adam S Aug 29 '14 at 01:44
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You can't break up the interval from $0 $ to $x$ like that because the integral of $1/x$ diverges, but you can try $\epsilon$ to $x$ for $\epsilon > 0$.

$$ \int_\epsilon^x \dfrac{1-\cos(t)}{t}\ dt + {\rm Ci}(x) = \ln(x) - \ln(\epsilon) - \int_{\epsilon}^\infty \dfrac{\cos(t)}{t}\ dt = \ln(x) - \ln(\epsilon) + {\rm Ci}(\epsilon)$$

So now what you need is to show that $$\lim_{\epsilon \to 0+} \left({\rm Ci(\epsilon)} - \ln(\epsilon)\right) = \gamma$$

Robert Israel
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  • I am weak in understanding in this type of problem solving. I rather stick with algebraic method. Nevertheless, I am grateful for all of your brainstorming. – Adam S Aug 29 '14 at 01:30