Find the principal value of :
$\displaystyle\int_{\mathbb{R}}\frac{e^{-x^3}}{x+3}dx$
$\displaystyle\int_{\mathbb{R}}\frac{e^{-x^2}}{x+1}dx$
Of course wolfram doesn't say convergence because problem in point $-3$
Definition of Cauchy principal value is :
$a≤t≤b$
$Pv\int_{[a,b]}f(x)dx$
$=\lim_{t\to 0^+}(\int_{[a,c-t]}f(x)dx+\int_{[c+t,b]}f(x)dx)$
But how I applied here?
Is contour integration work here?
The principal value is $-\infty$.
You have $$ P.V.\int_{-\infty}^\infty \frac{e^{-x^3}}{x+3}dx = \lim_{\epsilon\rightarrow 0_+, M\rightarrow\infty} \Big(\int_{-M}^{-3-\epsilon}\frac{e^{-x^3}}{x+3}dx + \int_{-3+\epsilon}^M\frac{e^{-x^3}}{x+3}dx \Big) = \\ = \lim_{\epsilon\rightarrow 0_+} \Big(\int_{-4}^{-3-\epsilon}\frac{e^{-x^3}}{x+3}dx + \int_{-3+\epsilon}^{-2}\frac{e^{-x^3}}{x+3}dx \Big) + \\ +\lim_{M\rightarrow \infty} \Big(\int_{-M}^{-4}\frac{e^{-x^3}}{x+3}dx + \int_{-2}^{M}\frac{e^{-x^3}}{x+3}dx \Big)$$
You can check that $\lim_{\epsilon\rightarrow 0_+}(\dots)$ and $ \lim_{M\rightarrow \infty} \int_{-2}^{M}\frac{e^{-x^3}}{x+3}dx $ are finite, but $$ \lim_{M\rightarrow \infty} \int_{-M}^{-4}\frac{e^{-x^3}}{x+3}dx = -\infty $$ so in total $$ P.V.\int_{-\infty}^\infty \frac{e^{-x^3}}{x+3}dx = -\infty $$