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Power series Calculator

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1

Solved example of power series

$\int\frac{\sin\left(x\right)}{x}dx$
2

Rewrite the function $\sin\left(x\right)$ as it's representation in Maclaurin series expansion

$\int\frac{\sum_{n=0}^{\infty } x^{\left(2n+1\right)}\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}}{x}dx$
3

Bring the denominator $x$ inside the power serie

$\int\sum_{n=0}^{\infty } \frac{x^{\left(2n+1\right)}{\left(-1\right)}^n}{x\left(2n+1\right)!}dx$
4

Simplify the fraction by $x$

$\int\sum_{n=0}^{\infty } \frac{x^{2n}{\left(-1\right)}^n}{\left(2n+1\right)!}dx$
5

We can rewrite the power series as the following

$\sum_{n=0}^{\infty } \frac{1}{\left(2n+1\right)!}\int x^{2n}{\left(-1\right)}^ndx$
6

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\int x^{2n}dx$
7

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\frac{x^{\left(2n+1\right)}}{2n+1}$
8

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\frac{x^{\left(2n+1\right)}}{2n+1}+C_0$

Final Answer

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\frac{x^{\left(2n+1\right)}}{2n+1}+C_0$

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