Using the power series expansions of e^x, sin x, cos x
Using the power series expansions of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^a$, expand $\displaystyle\int\frac{\cos x}x\,dx$ in a power series in $x$ and indicate when the expansion is valid.
Explanation of Method
Dividing the $\cos x$ series by $x$ isolates a $1/x$ term (whose antiderivative is $\ln|x|$) from an entire even-power remainder that can be integrated term by term without difficulty.
- Divide the Maclaurin series for $\cos x$ by $x$.
- Separate the $1/x$ term and integrate the rest term by term.
- Recognize the result as the cosine integral function $\operatorname{Ci}(x)$.
Step 1 · Divide by $x$
$$ \frac{\cos x}x = \frac1x\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n} = \frac1x + \sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n-1}. $$
Step 2 · Integrate
$$ \int\frac{\cos x}x\,dx = \ln|x| + \sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}\cdot\frac{x^{2n}}{2n} + C = \ln|x| + \sum_{n=1}^{\infty}\frac{(-1)^n}{2n\cdot(2n)!}x^{2n} + C. $$
Step 3 · Identify the function and check the derivative
This indefinite integral is (up to the choice of constant) the cosine integral function $\operatorname{Ci}(x)$. Differentiating back confirms the derivation: $$ \frac{d}{dx}\left[\ln|x|+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n}}{2n\cdot(2n)!}\right] = \frac1x+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n-1}}{(2n)!} = \frac1x+\frac{\cos x-1}x = \frac{\cos x}x. $$
Summary Table
| Property | Value |
|---|---|
| Series | $\ln|x|+\sum(-1)^nx^{2n}/(2n\cdot(2n)!)$ |
| Special function | Cosine integral, $\operatorname{Ci}(x)$ |
| Region of validity | $x\ne0$ |
Additional Notes
- The power-series part converges for all real $x$ (since $\cos x$ is entire), but the standalone $\ln|x|$ term introduces the only obstruction, a logarithmic singularity at $x=0$.
- This function cannot be expressed in terms of elementary functions — like the dilogarithm of Problem 264 and the exponential integral, it genuinely requires a new named special function, here the cosine integral.
- The construction closely parallels Problem 264's $\int\ln(1+t)/t\,dt$: both start by dividing a known Maclaurin series by the integration variable to expose (or, here, avoid) a $1/t$ singularity before integrating termwise.