How to calculate the integral $\int\limits_{-10}^{10}\frac{3^x}{3^{\lfloor x\rfloor}}dx$?

Question

I have to calculate this integral: $$\int\limits_{-10}^{10}\frac{3^x}{3^{\lfloor x\rfloor}}dx$$

I know that this function ie. $$3^{x-\lfloor x\rfloor}$$ is periodic with period $T=1$ so I rewrote the integral as $$20\int_{0}^{1}\frac{3^x}{3^{\lfloor x\rfloor}}dx$$

But the problem is that I can't figure out how to calculate the final integral.

Answer 1

Hint

$$\int_{-10}^{10} \frac{3^x}{3^{\lfloor x\rfloor }}\,\mathrm d x=\sum_{i=-9}^{10}\frac{1}{3^{k-1}}\int_{k-1}^k 3^x\,\mathrm d x.$$

Answer 2

$$\int_{-10}^{10}\frac{3^x}{3^{\lfloor x\rfloor}}dx=20\int_{0}^{1}\frac{3^x}{3^{\lfloor x\rfloor}}dx=20\int_{0}^{1}{3^x}dx=20\left[\frac{3^x}{\ln3}\right]^1_0=\frac{40}{\ln3}$$ For first step to second step, we note that $3^{\lfloor x\rfloor}=1$ for $0\le x<1$ as $\lfloor x\rfloor=0$.

Answer 3

$x-[x]=\{x\}$ is fractional part which is periodic with period 1. Your integral covers 20 cycles of period 1 So $$I=\int_{-10}^{10} 3^{\{x\}} dx =20 \int_{0}^{1} 3^x dx=20 \frac{3^1-1}{\ln3}= \frac{40}{\ln 3}.$$

Answer 4

$$20 \int\limits_{x=0}^1 3^x\ dx = \frac{40}{\ln 3}$$ (as given by Dr Azfar Ahmed)