Calculating a power series of a function

Question

I need to calculate the expression: $$ \int_0 ^1 \frac{1-\cos t}{t^2} \, dt $$ and given a hint: use the power series of $f(x)=\int_0 ^x \frac{\sin t}{t} \, dt $ and substitute $x=1$.

Now, I understand that: $2 \left( \frac{\sin t}{t }\right) ^2 =\frac{1-\cos t}{t^2} $ but how does this help me when using the hint? i.e. , if $2 \left( \frac{\sin t}{t }\right) ^2 =\frac{1-\cos t}{t^2} $ does not mean that $2 f(x)^2 = \int_0 ^1 \frac{1-\cos t}{t^2} dt $ , right?

Will you please help me figure out the "helpfulness" of the hint?

Thanks!

I would probably instead use the power series for $1-\cos t$, and divide term by term by $t^2$. Then integrate term by term. — André Nicolas, Nov 02 '15 at 19:10
It's not true that $2\sin^2 t = 1-\cos t$. Rather $2\sin^2 t = 1-\cos(2t)$. ${}\qquad{}$ — Michael Hardy, Nov 02 '15 at 20:05

score 1 · Accepted Answer · answered Nov 02 '15 at 19:25

Integration by parts gives:

$$ I=\int_{0}^{1}\frac{1-\cos t}{t^2}\,dt = -\left.\frac{1-\cos t}{t}\right|_{0}^{1}+\int_{0}^{1}\frac{\sin t}{t}\,dt\tag{1}$$ hence we have:

$$\begin{eqnarray*} I &=& -2\sin^2\left(\frac{1}{2}\right)+\int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n t^{2n}}{(2n+1)!}\,dt\\&=& -2\sin^2\left(\frac{1}{2}\right)+\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\cdot(2n+1)!}\\&=&\int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}t^{2n-2}}{(2n)!}\,dt\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n-1)\cdot(2n)!}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\cdot(2n+2)!}\\&=&-1+\cos(1)+\text{Si}(1).\tag{2}\end{eqnarray*}$$

Calculating a power series of a function

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