Get absolute value in exponent of integrand

Question

Let $C$ be some constant and $x\in\mathbb{R}$.

Can I write the expression $$ e^{-C\lvert x\rvert}\int_{-\infty}^\infty e^{-z^2+C\lvert x-z\rvert}\, dz $$ in the form (or can I estimate from above by) $$ e^{-C\lvert x\rvert}\int_{-\infty}^\infty e^{-z^2+C\lvert x\rvert+C\lvert z\rvert}\, dz? $$

I really would like to have the summand $C\lvert x\rvert$ in the exponent of the integrand, because I want to cancel the factor $e^{-C\lvert x\rvert}$ in front of the integral.

Answer 1

I would write

$\begin{array}\\ I(C, x) &=e^{-C\lvert x\rvert}\int_{-\infty}^\infty e^{-z^2+C\lvert x-z\rvert}\, dz\\ &=e^{-C\lvert x\rvert}\left(\int_{-\infty}^x e^{-z^2+C\lvert x-z\rvert}\, dz+\int_{x}^\infty e^{-z^2+C\lvert x-z\rvert}\, dz\right)\\ &=e^{-C\lvert x\rvert}\int_{-\infty}^x e^{-z^2+C (x-z)}\, dz+e^{-C\lvert x\rvert}\int_{x}^\infty e^{-z^2+C(z-x)}\, dz\\ &=e^{-C(\lvert x\rvert+x)}\int_{-\infty}^x e^{-z^2-Cz}\, dz+e^{-C(\lvert x\rvert-x)}\int_{x}^\infty e^{-z^2+Cz}\, dz\\ &=e^{-C(\lvert x\rvert+x)}\int_{-\infty}^x e^{-(z^2+Cz)}\, dz +e^{-C(\lvert x\rvert-x)}\int_{x}^\infty e^{-(z^2-Cz)}\, dz\\ &=e^{-C(\lvert x\rvert+x)}\int_{-\infty}^x e^{-(z^2+Cz+C^2/4)+C^2/4}\, dz +e^{-C(\lvert x\rvert-x)}\int_{x}^\infty e^{-(z^2-Cz+C^2/4)+C^2/4}\, dz\\ &=e^{-C(\lvert x\rvert+x)+C^2/4}\int_{-\infty}^x e^{-(z+C/2)^2}\, dz +e^{-C(\lvert x\rvert-x)+C^2/4}\int_{x}^\infty e^{-(z-C/2)^2}\, dz\\ \end{array} $

For any $x$, one of the exponents cancels depending on the sign of $x$.

Answer 2

Well, you could use the triangle inequality, $$ 0 \leq |x - z| \leq |x| + |z| $$ (on the right; the left is automatic from the definition of absolute value). If $C \geq 0$, $$ 0 \leq C|x - z| \leq C|x| + C|z| $$ and monotonicity gives , $$ \mathrm{e}^{-z^2} \leq \mathrm{e}^{-z^2 + C|x - z|} \leq \mathrm{e}^{-z^2 + C|x| + C|z|} \text{.} $$ However, if $C < 0$, $$ 0 \geq C|x - z| \geq C|x| + C|z| $$ and monotonicity gives , $$ \mathrm{e}^{-z^2} \geq \mathrm{e}^{-z^2 + C|x - z|} \geq \mathrm{e}^{-z^2 + C|x| + C|z|} \text{.} $$ So whether you get the kind of bound you say depends on the sign of $C$.

An entirely different way to go (that perhaps clarifies why the sign of $C$ matters): \begin{align*} &\int_{-\infty}^\infty \; \mathrm{e}^{-z^2 + C|x - z|} \,\mathrm{d}z \\ &= \int_{-\infty}^x \; \mathrm{e}^{-z^2 + C|x - z|} \,\mathrm{d}z + \int_{x}^\infty \; \mathrm{e}^{-z^2 + C|x - z|} \,\mathrm{d}z \\ &= \int_{-\infty}^x \; \mathrm{e}^{-z^2 + C(x - z)} \,\mathrm{d}z + \int_{x}^\infty \; \mathrm{e}^{-z^2 + C(z - x)} \,\mathrm{d}z \\ &= \mathrm{e}^{Cx} \int_{-\infty}^x \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{x}^\infty \; \mathrm{e}^{-z^2 + Cz} \,\mathrm{d}z \\ &= \mathrm{e}^{Cx} \int_{-\infty}^x \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z - \mathrm{e}^{-Cx}\int_{-x}^{-\infty} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u & \hspace{-.9in} \begin{bmatrix} u=-z \\ \mathrm{d}u = -\mathrm{d}z \end{bmatrix} \\ &= \mathrm{e}^{Cx} \int_{-\infty}^x \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-\infty}^{-x} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u \\ &= \begin{cases} \mathrm{e}^{Cx} \int_{-\infty}^x \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-\infty}^{-x} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u &, x \geq 0\\ \mathrm{e}^{Cx} \int_{-\infty}^x \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-\infty}^{-x} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u &, x < 0 \\ \end{cases} \\ &= \begin{cases} \mathrm{e}^{Cx} \int_{-\infty}^{-|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{Cx} \int_{-|x|}^{|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-\infty}^{-|x|} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u &, x \geq 0 \\ \mathrm{e}^{Cx} \int_{-\infty}^{-|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-\infty}^{-|x|} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u + \mathrm{e}^{-Cx}\int_{-|x|}^{|x|} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u &, x < 0 \\ \end{cases} \\ &= \begin{cases} \left( \mathrm{e}^{Cx} + \mathrm{e}^{-Cx} \right) \int_{-\infty}^{-|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{Cx} \int_{-|x|}^{|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z &, x \geq 0 \\ \left( \mathrm{e}^{Cx} + \mathrm{e}^{-Cx} \right) \int_{-\infty}^{-|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{-Cx}\int_{-|x|}^{|x|} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u &, x < 0 \\ \end{cases} \\ &= \left(\mathrm{e}^{C|x|} + \mathrm{e}^{-C|x|}\right) \int_{-\infty}^{-|x|} \; \mathrm{e}^{-z^2 - Cz} \,\mathrm{d}z + \mathrm{e}^{C|x|}\int_{-|x|}^{|x|} \; \mathrm{e}^{-u^2 - Cu} \,\mathrm{d}u \text{.} \end{align*} After multiplying through by your pre-factor, exactly the cancellation you want will occur on the right term and will partially occur in the left term.