I'm trying to figure out how to evaluate the integral
$\int_{-\infty}^{\infty}\frac{1}{x^2+t^2}\exp^{-\frac{x^2}{2t}}dx$
where $t\geq 0$.
I have tried a lot of methods but without success. Please help.
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1This doesn't have a closed form unless you take the error function as an elementary function. – Peter Foreman Apr 21 '19 at 21:48
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How can be solved in that case? – knot killer Apr 21 '19 at 21:57
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Given the convergence condition $\Re t>0$, the result (in terms of the complementary error function $\operatorname{erfc}x:=\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt$) is $\frac{\pi}{t}e^{t/2}\operatorname{erfc}\sqrt{\frac{t}{2}}$ according to Wolfram Alpha. @RobinNicole's proof strategy gets this, if you know the error function's series representation. – J.G. Apr 21 '19 at 22:10
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1You should use $1/(t^2+ x^2) = 1/t^2 \sum_n (-1)^n (x/t)^{2 n}$ and then by looking for the formula of the odds moments of a normal distribution you shoudl be able to make the calculations – Robin Nicole Apr 21 '19 at 22:11
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Change variables in your integral, letting $x = ty$. Then your integral becomes $${1 \over t} \int_{-\infty}^{\infty}\frac{1}{y^2 + 1}e^{-\frac{ty^2}{2}}dy$$ Thus you are looking for ${\displaystyle {1 \over t}f\bigg({t \over 2}\bigg)}$, where $$f(u) = \int_{-\infty}^{\infty}\frac{1}{y^2 + 1}e^{-uy^2}dy$$ By differentiating under the integral sign, you have $$f'(u) - f(u) = -\int_{-\infty}^{\infty}e^{-uy^2}dy$$ By changing variables $z = u^{1/2}y$ this is equal to $$-u^{-{1 \over 2}}\int_{-\infty}^{\infty}e^{-z^2}dz$$ $$= -\sqrt{\pi}u^{-{1 \over 2}}$$ So $f(u)$ satisfies the differential equation $$f'(u) - f(u) = -\sqrt{\pi}u^{-{1 \over 2}}$$ $$f(0) = \pi$$ Now we can use standard first order linear methods, using an integrating factor of $e^{-u}$ to get $$(e^{-u} f(u))' = - \sqrt{\pi}u^{-{1 \over 2}}e^{-u}$$ Hence $f(u)$ is given by $$-\sqrt{\pi}e^{u} (\int^u v^{-{1 \over 2}}e^{-v}\,dv + C)$$ Changing variables $v = w^2$ leads to $$f(u) = -2\sqrt{\pi}e^{u} (\int^{u^{1 \over 2}} e^{-w^2}\,dw + C)$$ $$= -\pi e^{u}(erf(u^{1 \over 2}) + C)$$ Now it just becomes a question of plugging in $u = 0$ to get $C$, and then letting ${1 \over t}f(t/2)$ be your final answer. Plugging in $u = 0$ gives $C = -1$, so that $$f(u) = -\pi e^{u}(\rm{erf}(u^{1 \over 2}) - 1)$$ $$= \pi e^u {\rm erfc} (u^{1 \over 2})$$ So your final answer is just $$ {\pi \over t}e^{t \over 2}{\rm erfc}\bigg(\sqrt{t \over 2}\bigg)$$
Zarrax
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