# Please how solve this Improper Integral

Please how solve this Improper Integral ... $$\int_{1}^{+\infty}\frac{1}{\sqrt[x]{e}}dx$$... i know how to solve $$\int_1^{+\infty} \sqrt[x]{e}$$ I create a Taylor Polynomial and integrate, but this is a little hard. if it is possible...

Presumably you were able to show that $\int_1^\infty \sqrt[x]{e}\,dx$ diverges, because $$\lim_{x\to\infty} \sqrt[x]{e} = 1$$ But it follows from the above that $$\lim_{x\to\infty} \frac{1}{\sqrt[x]{e}} = \frac{1}{1} = 1$$ so $\int_1^\infty \frac{1}{\sqrt[x]{e}}\,dx$ diverges, too.
Through the substitution $x=\frac{1}{z}$ both integrals boil down to $$\int_{0}^{1}\frac{\exp(\pm z)}{z^2} \,dz$$ that is a diverging integral due to the non-integrable singularity at the origin.
The derivative of $$e^{-\frac{1}{x}}$$ is such that $$\frac{d}{dx}e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x^2}>0$$ so it is an increasing function, thus it diverges.
Thanks to all, but I think I've got the answer. $$\int \frac{1}{\sqrt[x]{e}} dx = \int \frac{e^{-\frac{1}{x}}}{x^2}x^2dx$$ so I can solve this by parts....$$dv=\frac{e^{-\frac{1}{x}}}{x^2}dx$$ so $$v=e^{-\frac{1}{x}}$$ and $$u=x^2$$ makes $$du=2xdx$$
In that case my integral now is $$\int e^{-\frac{1}{x}}dx =uv-\int vdu=x^2e^{-\frac{1}{x}}-2\int xe^{-\frac{1}{x}}dx$$ And again, applying the same idea to $$\int xe^{-\frac{1}{x}}dx=\int\frac{e^{-\frac{1}{x}}}{x^2}x^3dx$$