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Here is one solution. Let

$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$

Clearly,

$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$

We assume that $x\ne 0$. Then by change of variable

$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$

the integral becomes

$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:,$$

where we have applied the integral (7.4.9) in P. 302

$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a frac{2\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$

with $a=1$ and $z=\frac{x^2}{2\sigma t}$. , where we have used the fact that $\text{Erfc}(x)=2(1-\Phi(x))$. This then proves

$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:.$$

Finally, the case that $x=0$ can be easily verified. This then finishes the proof.

3 added 2 characters in body

Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)

Here is one solution. Let

$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$

Clearly,

$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$ Then by change of variable

$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$

the integral becomes

$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:,$$

where we have applied the integral (7.4.9) in P. 302

$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$

with $a=1$ and $z=\frac{x^2}{2\sigma t}$. This then proves

$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:.$$

2 added 21 characters in body

Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)

Here is one solution. Let

$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$

Clearly,

$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$ Then by change of variable

$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$

the integral becomes

$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma t}} \right)\right)\:,$$

where we have applied the integral (7.4.9) in P. 302

$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\:$$

with $a=1$ and $z=\frac{x^2}{2\sigma t}$. This then proves

$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{\sigma t}} \right)\right)}\:.$$

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