Introduction

Happy new year blog

On the last day of 2019 Lets solve some problems so that we prepare for 2019.

Did you know that there is no explicit formula for the indefinite integral:

$\int \; <!--\; integral\; -->e^{-x^2}\mathrm{dx}=?$

But we have a value
for definite integral

$\underset{0}{\overset{\mathrm{\infty \; <!--\; infinity\; -->}}{\int \; <!--\; integral\; -->}}{e}^{-x^2}\mathrm{dx}=\frac{\sqrt{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}}{2}$

Do you use the symmetry while solving definite integrals? That is a useful trick; any function can be written as

$g\left(x\right)=g_e\left(x\right)+g_0\left(x\right)$

$\begin{array}{}g\left(-x\right)=g_e\left(-x\right)+g_0\left(-x\right)=g_e\left(x\right)-g_0\left(x\right)\\ g_e\left(x\right)=\frac{\left[g\left(x\right)+g\left(-x\right)\right]}{2}\\ g_0\left(x\right)=\frac{\left[g\left(x\right)-g\left(-x\right)\right]}{2}\end{array}$

So that;

Now lets find integral;

$\underset{-1}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{cos\left(x\right)}{e^{\frac{1}{x}}+1}=?$

$g_e\left(x\right)=\frac{cos\left(x\right)}{2}\times \; <!--\; multiplication\; sign\; -->\left(\frac{1}{e^{1/x}+1}+\frac{1}{e^{-1/x}-1}\right)=\frac{cos\left(x\right)}{2}$

Now, odd part integral is zero since interval is symmetric.

$\underset{-1}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{cos\left(x\right)}{e^{1/x}+1}\mathrm{dx}=\underset{-1}{\overset{1}{\int \; <!--\; integral\; -->}}{g}_e\left(x\right)\mathrm{dx}=\underset{-1}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{cos\left(x\right)}{2}\mathrm{dx}=sin\left(1\right)$