Definite integral of $e^{-by^2}/(y^2-c^2)$

Question

How to get the definite integral like this?

$$\int^a_0 \frac{\exp(-by^2)}{y^2-c^2}dy$$ where $a,b,c$ are parameters and $a>c$.

Thanks a lot.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\tt\large\mbox{Let's assume}\ \verts{c} > a > 0$:

$\ds{I \equiv \int^{a}_{0}{\expo{-by^{2}} \over y^{2} - c^{2}}\,\dd y = {1 \over \verts{c}}\int^{a/\verts{c}}_{0} {\expo{-bc^{2}y^{2}} \over y^{2} - 1}\,\dd y = {1 \over \verts{c}}\,{\cal F}\pars{bc^{2},{a \over \verts{c}}}}$

where $\ds{{\cal F}\pars{\mu,\nu} \equiv \int_{0}^{\nu}{\expo{-\mu x^{2}} \over x^{2} - 1}\,\dd x}$

\begin{align} &\partiald{{\cal F}\pars{\mu,\nu}}{\mu} = \int_{0}^{\nu}{\expo{-\mu x^{2}}\pars{-x^{2}} \over x^{2} - 1}\,\dd x = -\int_{0}^{\nu}\expo{-\mu x^{2}}\,\dd x - \int_{0}^{\nu}{\expo{-\mu x^{2}} \over x^{2} - 1}\,\dd x \\[3mm]&= -\,{1 \over 2}\root{\pi \over \mu}\,{2 \over \root{\pi}} \int_{0}^{\nu\root{\mu}}\expo{x^{2}}\,\dd x - {\cal F}\pars{\mu,\nu} = -\,{1 \over 2}\root{\pi \over \mu}\,{\rm erf}\pars{\nu\root{\mu}} - {\cal F}\pars{\mu,\nu} \\[3mm]& \pars{\partiald{}{\mu} + 1}{\cal F}\pars{\mu,\nu} = -\,{1 \over 2}\root{\pi \over \mu}\,{\rm erf}\pars{\nu\root{\mu}} \\[3mm]& \partiald{\bracks{\expo{\mu}{\cal F}\pars{\mu,\nu}}}{\mu} = -\,{1 \over 2}\,\expo{\mu}\root{\pi \over \mu}\,{\rm erf}\pars{\nu\root{\mu}} \\[3mm]& \expo{\mu}{\cal F}\pars{\mu,\nu} - {\cal F}\pars{0,\nu} = -\,{\root{\pi} \over 2} \int_{0}^{\mu}{\expo{\mu'}{\rm erf}\pars{\nu\root{\mu'}} \over \root{\mu'}} \,\dd \mu' \\[3mm]& {\cal F}\pars{\mu,\nu} = \expo{-\mu}{\cal F}\pars{0,\nu} - \expo{-\mu}\,{\root{\pi} \over 2} \int_{0}^{\mu}{\expo{\mu'}{\rm erf}\pars{\nu\root{\mu'}} \over \root{\mu'}} \,\dd \mu' \end{align}

$$ {\cal F}\pars{0,\nu} = \int_{0}^{\nu}{\dd x \over x^{2} - 1}\,\dd x = {1 \over 2}\int_{0}^{\nu}\pars{{1 \over x - 1} - {1 \over x + 1}}\,\dd x = {1 \over 2}\ln\pars{1 - \mu \over \mu + 1} $$ $$\color{#0000ff}{\large% {\cal F}\pars{\mu,\nu} = {1 \over 2}\,\expo{-\mu}\ln\pars{1 - \mu \over \mu + 1} - \expo{-\mu}\,{\root{\pi} \over 2} \int_{0}^{\mu}{\expo{\mu'}{\rm erf}\pars{\nu\root{\mu'}} \over \root{\mu'}} \,\dd \mu'} $$

Answer 2

If $0<c<a$, then this integral is improper at $y=c$. So, we must write $$ \int_0^a\frac{e^{-by^2}}{y^2-c^2}\,dy=\int_0^c\frac{e^{-by^2}}{y^2-c^2}\,dy+\int_c^a\frac{e^{-by^2}}{y^2-c^2}\,dy, $$ and check whether these two integrals each converge.

For the first, note that for $y\in(0,c)$, $e^{-bc^2}\leq e^{-by^2}\leq 1$. Since $y^2-c^2<0$, it follows that $$ \frac{1}{y^2-c^2}\leq \frac{e^{-by^2}}{y^2-c^2}\leq\frac{e^{-bc^2}}{y^2-c^2} $$ for all $y\in(0,c)$.

From here, it is not difficult to check that $$ \lim_{M\rightarrow c^{-}}\int_0^M\frac{e^{-bc^2}}{y^2-c^2}\,dy=-\infty, $$ so that the integral you want diverges by comparison.

So, since one of the parts of your integral diverges, the whole integral diverges.