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Reading the defintion of the IntegralCosinus $$ {\rm Ci}(x) = \gamma + \ln x + \int_0^x\frac{\cos t-1}{t}\,dt $$ I wonder what happens, if I to split the function in the integral: $$ \begin{eqnarray*} {\rm Ci}(x) &=& \gamma + \ln x + \int_0^x\frac{\cos t}{t}\,dt -\int_0^x\frac{1}{t}\,dt \\ &=& \gamma + \ln x + \int_0^x\frac{\cos t}{t}\,dt -\left[ \ln (t) \right]_0^x \\ &=& \gamma + \ln x + \int_0^x\frac{\cos t}{t}\,dt -\ln(x) + \underbrace{\ln(0)}_? \tag{1} \\ \end{eqnarray*} $$ Is splitting not allowed here or how do I have to interprete $\ln(0)$?

And further if I look at another definition $$ -{\rm Ci}(x) = \int_x^\infty\frac{\cos t}{t}\,dt \tag{2} $$ and now add $(1)$ and $(2)$ I get: $$ 0=\gamma + \int_0^\infty\frac{\cos t}{t}\,dt $$ This doesn't seem right. Can anybody tell me what's wrong here?

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3 Answers 3

up vote 7 down vote accepted

This is questionable: $$ \int_0^x\frac{\cos t - 1}{t}\,dt = \int_0^x\frac{\cos t}{t}\,dt -\int_0^x\frac{1}{t}\,dt $$ A convergent integral written as the difference of two divergent integrals.

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In the first definition, $\log(x)$ is not defined for $x = 0$, and in the second definition, the integral is not defined when $x = 0$, so adding the two results together is not a valid operation when $x = 0$.

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What if I assume $x>0$? The definitions would stay the same. –  draks ... Jan 21 '12 at 22:25

Notice that $$ \frac{\cos t}{t} \to \infty\text{ as }t \downarrow 0 $$ and $$ \frac 1t \to \infty\text{ as }t\downarrow 0 $$ but $$ \frac{\cos t - 1}{t} \to 0\text{ as }t\downarrow 0. $$ So it makes sense to speak of the integral of this last fraction from $0$ to $x$. What about the first two? Notice that for $x>0$, $$ \int_0^x \frac 1t\;dt = \infty $$ and $$ \int_0^x \frac{\cos t}{t}\;dt \ge \int_0^{\cos x} \frac 1t\;dt = \infty. $$ But the integral of the difference is the integral of a bounded function, so it's well-behaved.

(But don't conclude that $\infty-\infty=0$. There are instances of two functions approaching $\infty$ while their difference approaches $6$. And similarly with any other number in place of $6$.)

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