Sometimes just using the integral form of error function can save you.
$\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}
\text{erf}\biggl(\dfrac{b}{\sqrt{x}}\biggr)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^\frac{b}{\sqrt{x}}e^{-u^2}~du~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1e^{-\left(\frac{bu}{\sqrt{x}}\right)^2}~d\biggl(\dfrac{bu}{\sqrt{x}}\biggr)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1\dfrac{b}{\sqrt{x}}e^{-\frac{b^2u^2}{x}}~du~dx$
$=\dfrac{2b}{\sqrt{\pi}}\int_0^t\int_0^1\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}du~dx$
$=\dfrac{2b}{\sqrt{\pi}}\int_0^1\int_0^t\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^t-e^{-\frac{b^2u^2+a^2}{x}}~d\biggl(\dfrac{1}{\sqrt{x}}\biggr)~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\infty^\frac{1}{\sqrt{t}}-e^{-(b^2u^2+a^2)x^2}~dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\frac{1}{\sqrt{t}}^\infty e^{-(b^2u^2+a^2)x^2}~dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\infty e^{-(b^2u^2+a^2)x^2}~dx~du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\frac{1}{\sqrt{t}}e^{-(b^2u^2+a^2)x^2}~dx~du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^nx^{2n+1}}{n!(2n+1)}\biggr]_0^\frac{1}{\sqrt{t}}~du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}b^{2k}u^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}du$
$=2b\biggl[\dfrac{\ln\left(b^2u+b\sqrt{b^2u^2+a^2}\right)}{b}\biggr]_0^1-\dfrac{4b}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}b^{2k}u^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^1$
$=2\ln\left(b^2+b\sqrt{a^2+b^2}\right)-2\ln(|a|b)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n4a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$
$\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\text{erfc}\left(\sqrt{\dfrac{a^2+x^2}{t}}\right)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_\sqrt{\frac{a^2+x^2}{t}}^\infty e^{-u^2}~du~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\infty e^{-u^2}~du~dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\sqrt{\frac{a^2+x^2}{t}}e^{-u^2}~du~dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^nu^{2n+1}}{n!(2n+1)}\biggr]_0^\sqrt{\frac{a^2+x^2}{t}}~dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^{n+\frac{1}{2}}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}x^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\left[\ln\left(x+\sqrt{a^2+x^2}\right)\right]_0^b-\dfrac{2}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}x^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^b$
$=\ln\left(b+\sqrt{a^2+b^2}\right)-\ln|a|-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$
First, note that $\sin^2(tx)=\frac12(1-\cos(2tx))$. Hence, we see that
$$I(t)=\frac\pi4-\frac12 \int_0^\infty \frac{\cos(2tx)}{x^2+1}\,dx\tag1$$
Differentiating under the integral in $(1)$ can be justified by noting that the integral $\int_0^\infty \frac{x\sin(2tx)}{x^2+1}\,dx$ converges uniformly for $|t|\ge \delta>0$. Proceeding reveals
$$\begin{align}
I'(t)&=\int_0^\infty \frac{x\sin(2tx)}{x^2+1}\,dx\\\\
&=\int_0^\infty \frac{(x^2+1-1)\sin(2tx)}{x(x^2+1)}\,dx\\\\
&=\int_0^\infty \frac{\sin(2tx)}{x}\,dx-\int_0^\infty \frac{\sin(2tx)}{x(x^2+1)}\,dx\\\\
&=\frac\pi2 \text{sgn}(t)-\int_0^\infty \frac{\sin(2tx)}{x(x^2+1)}\,dx\tag2
\end{align}$$
Similarly, we can differentiate $(2)$ to obtain
$$\begin{align}
I''(t)&=-2\int_0^\infty \frac{\cos(2tx)}{x^2+1}\,dx\\\\
&=4I(t)-\pi\tag3
\end{align}$$
From $(3)$ we have $I''(t)-4I(t)=-\pi$, while from $(1)$ we see that $I(0)=0$ and from $(2)$ we see that $\lim_{t\to 0^\pm}I'(t)=\pm \frac\pi2$. Solving this ODE with these initial conditions, we find
$$I(t)=\frac\pi4 -\frac\pi4 e^{-2|t|}$$
Best Answer
Define a function $g$ by $$g(n,x,t)=\frac{\sin(x n)}{xn} e^{-t n^2}$$ for $n,x,t > 0$. Now, $$\frac{\partial g}{\partial t}(n,x,t)=-n\frac{\sin(x n)}{x} e^{-t n^2}$$ Therefore $$\int_0^\infty \frac{\partial g}{\partial t}(n,x,t) dn = -\frac1{2x}\int_0^\infty \sin(nx)e^{-tn^2} 2ndn \\=-\frac1{2x}\int_0^\infty \sin(\sqrt{n}x)e^{-tn} dn$$ By the Laplace transform of $\sin(\sqrt{n}x)$, we have $$\frac1{x}\mathcal{L}\{\sin(\sqrt{n}x)\}(t)=\frac1{x}\int_0^\infty \sin(\sqrt{n}x)e^{-tn} dn=\frac{\sqrt{\pi}e^{-x^2/4t}}{2t^{\frac32}}$$ Now since $$-\frac{\partial }{\partial t}\int_0^\infty \frac{\sin(x n)}{xn} e^{-t n^2} dn =\frac{\sqrt{\pi}e^{-x^2/4t}}{4t^{\frac32}} $$ you can get the result finally beacuse $$\frac{\partial }{\partial t}\text{erf} \left(\frac{x}{2 \sqrt{t}}\right)=-\frac{x e^{-x^2/4t}}{2\sqrt{\pi}t^{\frac32}}$$ and $$\lim_{t\to\infty}\text{erf} \left(\frac{x}{2 \sqrt{t}}\right)=\text{erf}(0)=0\text{ for all }x>0$$