I want to compute the integral
$$
\int_t^T \int_{-\infty}^\infty \frac{1}{ \sqrt{\tau – t} (T-\tau)} \exp\left(-\frac{(z-x)^2}{2(\tau – t)} -\frac{(z-v)^2 + (z-w)^2}{2(T-\tau)} \right) d z d \tau.
$$
First integrating with respect to $z$ I get
$$
\int_t^T \frac{\sqrt{2\pi}}{ \sqrt{T – \tau}\sqrt{-2 t + \tau + T}}\exp\left(-\frac{(v + w – 2 x)^2}{4 (-2 t + \tau + T)} – \frac{(v-w)^2}{4 (T-\tau )}\right) d \tau.
$$
So to go further I think I should be able to integrate
$$
\int_0^T \frac{1}{ \sqrt{T^2 – \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} – \frac{b^2}{4 (T-\tau )}\right) d \tau
$$
Integrate $\frac{1}{ \sqrt{T^2 – \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} – \frac{b^2}{4 (T-\tau )}\right)$
definite integralsintegration
Related Solutions
The integral is somehow nontrivial. I would proceed as follows.
OP's second attempt seems to be on the right track (at least for my approach). Denote the integral as $I$, substitute $\dfrac{x^4-1}{2x^2}=t$ so that $$ I=\int_0^\infty\frac{dx}{\sqrt{1+\exp\left(\dfrac\pi2\left(x^2-\dfrac1{x^2}\right)\right) }}=\frac12\int_{-\infty}^\infty\frac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}}\frac{dt}{\sqrt{e^{\pi t}+1}} $$ The term $\dfrac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}}$ reminds me of $\sin(\theta/2)$, so I came up with $$ \int_{-\infty }^{\infty } \frac{1}{\left(v^2-t\right)^2+1} dv=\Im \int_{-\infty }^{\infty } \frac{1}{v^2-t-i} dv =\Im\frac{\pi}{\sqrt{-t-i}} =\frac\pi{\sqrt2}\dfrac{\sqrt{\sqrt{t^2+1}+t}}{ \sqrt{t^2+1}} $$ and the integral becomes $$ I=\frac1{\sqrt2\pi}\int_{-\infty }^{\infty }\int_{-\infty}^\infty \frac{1}{\left(v^2-t\right)^2+1}\frac{1}{\sqrt{e^{\pi t}+1}}dvdt\\ =\frac1{\sqrt2\pi}\int_{-\infty }^{\infty }du\int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt \tag{$\ast$} $$ where $v=u\sqrt{e^{\pi t}+1}$.
Now consider the meromorphic function $f(z)=\dfrac1{u^2(e^{\pi z}-1)+z}$ and integrate on the rectangular contour $[-R,R]\times [-1,1]$. When $R\to\infty$, it is easy to see that the integrals on the vertical lines tend to vanish as $e^{-\pi R},\Re(z)=R\to\infty$ and $R^{-1}, \Re(z)=-R\to -\infty$. Meanwhile, the two on the horizontal line combines to get $$ \oint f(z)dz=\int_{-\infty}^\infty f(t-i)dt+\int^{-\infty}_\infty f(t+i)dt=2i\int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt $$ Thus, it suffice to investigate the location of the poles of $f(z)$. Though it normally has infinite poles on the complex plane, I claim that there's only one in the contour, namely $z=0$. To see this, let $z=\Re(z)+i \Im(z)=\xi+i\eta$ be a pole, then take the real and imaginary part $$ \cases{ u^2(e^{\pi\xi}\cos(\pi\eta)-1)+\xi=0 &(r)\\[5pt] u^2e^{\pi \xi}\sin(\pi \eta)+\eta=0 &(i) } $$ The two terms on the LHS of equation (i) always has the same sign when $-\pi<\eta<\pi$, so its only solution in the contour is $\eta=0$. Now equation (r) becomes $u^2(e^{\pi\xi}-1)+\xi=0$. It is easy to see that LHS is a monotonic function of $\xi$ and has a root $\xi=0$, so it is the only one, proving the claim.
By the residue theorem, $$ \oint f(z)dz=2\pi i\text{Res}[f(z),0]=\frac{2\pi i}{1+\pi u^2} $$ that is $$ \int_{-\infty}^\infty \frac{1}{\left(u^2(e^{\pi t}+1)-t\right)^2+1}dt=\frac{\pi}{1+\pi u^2} $$ plug this into $(\ast)$ $$ I=\frac1{\sqrt2\pi}\int_{-\infty}^\infty \frac{\pi}{1+\pi u^2}du=\sqrt{\frac{\pi}2} $$ as desired.
Update
I played around with my method and found more interesting result as one can plug any analytic function into $f(z)$ that keeps the contour integral neet. For example, $$ \int_0^\infty \frac{dx}{\sqrt{\cosh(x\sqrt{x^2+\pi})}}=\frac{\pi}{2\sqrt2} $$ Denote $t(x)=\dfrac{x^4-1}{2x^2}$ and $$ J_n=\int_0^\infty\frac{dx}{\sqrt{1+e^{\pi t(x)}}^{2n+1}}=\sqrt{\frac{\pi }{2}}\cdot[z^{-1}]\frac{1}{\sqrt{z \left(e^{ z}-1\right)} \left(1-e^{ z}\right)^n} $$ is always an rational multiple of $\sqrt{2\pi}$, like $$ J_1=\frac{3 }{8}\sqrt{2\pi}\quad J_2=\frac{65}{192}\sqrt{2\pi}\quad J_3=\frac{245 }{768}\sqrt{2\pi}\quad \cdots $$ One with more poles in the contour $$ \int_0^\infty\frac{\operatorname{sech}(\pi t(x))}{\sqrt{1+e^{\pi t(x)}}}dx=\sqrt{\sqrt{2}+1}-\sqrt{\frac{\pi }{2}} $$
However, now some of them seem to be trivial under @Sangchul Lee's brilliant approach.
Your computations seem correct; however, you didn't choose the easiest path to handle this problem. Indeed, setting $f_c(t) := t^{-3/2}e^{-c/t}$, we may recast the considered integral as a (truncated) convolution product, as follows : $$ \int_0^T \frac{e^{ia/(T-\tau)}}{(T-\tau)^{3/2}} \frac{e^{ib/\tau}}{\tau^{3/2}} \mathrm{d}\tau = (f_{c_1}*f_{c_2})(T), $$ with $c_1 = -ia$ and $c_2 = -ib$. Let's recall that is the convolution product is mapped to the usual product through the Laplace transform. Since the Laplace transform of $f_c$ is given by $F_c(s) = \sqrt{\frac{\pi}{c}} e^{\sqrt{-4cs}}$ (see Gradstein & Ryzhik for instance), one has thus : $$ \begin{align} \mathscr{L}[f_{c_1}*f_{c_2}](s) &= F_{c_1}(s)F_{c_2}(s) \\ &= \frac{\pi}{\sqrt{c_1c_2}} e^{\sqrt{-4c_1s}+\sqrt{-4c_2s}} \\ &= \frac{\pi}{\sqrt{c_1c_2}} \exp\left(\sqrt{-4s(\sqrt{c_1}+\sqrt{c_2})^2}\right) \\ &= \sqrt{\pi}\frac{\sqrt{c_1}+\sqrt{c_2}}{\sqrt{c_1c_2}} F_{(\sqrt{c_1}+\sqrt{c_2})^2}(s) \end{align} $$ As the result displays the same structure, the inverse transform can be deduced straightforwardly, namely $\mathscr{L}^{-1}[F_{(\sqrt{c_1}+\sqrt{c_2})^2}](T) = f_{(\sqrt{c_1}+\sqrt{c_2})^2}(T)$, hence the desired result (up to the $-i$ factors) : $$ \int_0^T \frac{e^{ia/(T-\tau)}}{(T-\tau)^{3/2}} \frac{e^{ib/\tau}}{\tau^{3/2}} \mathrm{d}\tau = \sqrt{\pi}\frac{\sqrt{c_1}+\sqrt{c_2}}{\sqrt{c_1c_2}} \frac{e^{-(\sqrt{c_1}+\sqrt{c_2})^2/T}}{T^{3/2}} $$
Final remark. Even if this relation may seem arbitrary at first glance and result from the "luck of computations", it comes from a deeper property actually. Indeed, the function $f_c(t)$ corresponds to the density of a Lévy distribution. Moreover, the convolution of two PDFs or equivalently the product of their characteristic functions can be interpreted as the distribution of their sum. Yet, the Lévy distribution is a stable law, meaning that linear combinations of Lévy-distributed random variables will be Lévy-distributed too (with other parameters, nonetheless), that is why the same structure was preserved through the product of Laplace transforms.
Edit (see comments). Coming back to the initial parametrization, we get : $$ \begin{align} \int_0^T \frac{e^{ia/(T-\tau)}}{(T-\tau)^{3/2}} \frac{e^{ib/\tau}}{\tau^{3/2}} \mathrm{d}\tau &= \sqrt{\pi}\frac{\sqrt{-ia}+\sqrt{-ib}}{\sqrt{-ab}} \frac{e^{-(\sqrt{-ia}+\sqrt{-ib})^2/T}}{T^{3/2}} \\ &= \sqrt{\pi}\sqrt{\frac{-i}{-1}}\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}} \frac{e^{-\left(\sqrt{-i}(\sqrt{b}+\sqrt{b})\right)^2/T}}{T^{3/2}} \\ &= \sqrt{\frac{i\pi}{T^3}} \frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}} \exp\left(\frac{i}{T}\left(\sqrt{a}+\sqrt{b}\right)^2\right) \end{align} $$
Best Answer
$$I=\underbrace{\int_0^{T}\frac{d\tau}{\sqrt{T^2-\tau^2}}\exp\left(-\frac{a^2}{4(T+\tau)}-\frac{b^2}{4(T-\tau)}\right)}\\{\tau\rightarrow \frac{T(1-y^2)}{1+y^2}}$$
$$=\exp\left(-\frac{a^2+b^2}{8T}\right)\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\frac{a^2y^2+\frac{b^2}{y^2}}{8T}\right)$$
Let's parametrize the integral in order to build two ODE: $$ f(\alpha,\beta)=\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)$$
$$g(\alpha,\beta)=f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=\int_{0}^{1}\exp\left(-\alpha^2y^2-\frac{\beta^2}{y^2}\right)dy$$
Now, let's use $g(\alpha,\beta)$ to solve the remaining integral: $$\alpha g(\alpha,\beta)+\frac{1}{2}\frac{\partial g}{\partial\beta}=\exp\left(2\alpha\beta\right)\int_{0}^{1}\left(\alpha-\frac{\beta}{y^2}\right)\exp\left[-\left(\alpha y+\frac{\beta}{y}\right)^2 \right]dy=-\frac{\sqrt{\pi}}{2}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\frac{\partial g}{\partial\beta}+2\alpha g(\alpha,\beta)=-\sqrt{\pi}\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\mu_1(\beta)=\exp\left(\int 2\alpha d\beta\right)=\exp\left(2\alpha\beta\right)$$
$$\exp\left(2\alpha\beta\right)\frac{\partial g}{\partial\beta}+2\alpha\exp\left(2\alpha\beta\right) g(\alpha,\beta)=-\sqrt{\pi}\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)$$
$$\left(\exp\left(2\alpha\beta\right)g(\alpha,\beta)\right)'=-\sqrt{\pi}\int \exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)d\beta=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(4\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\operatorname{erf}\left(\alpha-\beta\right)\right)$$
$$g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)+c$$
$$g\left(\alpha,0\right)=\int_{0}^{1}\exp\left[-\left(\alpha y\right)^2 \right]dy=\frac{\sqrt{\pi}}{2\alpha}\ \operatorname{erf}\left(\alpha\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\ \operatorname{erfc}\left(\alpha\right)-\ \operatorname{erf}\left(\alpha\right)\right)+c\\ c=\frac{\sqrt{\pi}}{4\alpha}$$
$$\boxed{g\left(\alpha,\beta\right)=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)}$$
Using this result in the first ODE: $$f(\alpha ,\beta)-\frac{1}{2\alpha}\frac{\partial f}{\partial\alpha}=-\frac{\sqrt{\pi}}{4\alpha}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$
$$\frac{\partial f}{\partial\alpha}-2\alpha f(\alpha ,\beta)=\frac{\sqrt{\pi}}{2}\left(\exp\left(2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-1\right)$$
$$\mu_2(\alpha)=\exp\left(-\int 2\alpha d\alpha\right)=\exp\left(-\alpha^2\right)$$
$$\left(\exp\left(-\alpha^2\right)f(\alpha,\beta)\right)'=\frac{\sqrt{\pi}}{2}\int\left(\exp\left(-\alpha^2+2\alpha\beta\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\exp\left(-\alpha^2-2\alpha\beta\right)\ \operatorname{erf}\left(\alpha-\beta\right)-\exp\left(-\alpha^2\right)\right)d\alpha$$
$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\int\left(\underbrace{\exp\left(-\left(\alpha-\beta\right)^2\right)\ \operatorname{erfc}\left(\alpha+\beta\right)}_{IBP}-\exp\left(-\left(\alpha+\beta\right)^2\right)\ \operatorname{erf}\left(\alpha-\beta\right)\right)d\alpha$$
$$=-\frac{\operatorname{erf}\left(\alpha\right)\sqrt{\pi}}{2}+\frac{e\sqrt{\pi}}{2}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)-\int\left(\underbrace{\operatorname{erf}\left(\beta-\alpha\right)+\operatorname{erf}\left(\beta+\alpha\right)}_{0}\right)\exp\left(-(\alpha+\beta)^2\right)d\alpha\right)$$
$$f(\alpha ,\beta)=-\frac{\sqrt{\pi}\operatorname{erf}\left(\alpha\right)e^{\alpha^2}}{2}-\frac{\pi e^{a^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+k$$
$$f(0,0)=\int_0^1\frac{dy}{1+y^2}=\frac{\pi}{4}=k$$
$$\boxed{f(\alpha,\beta)=-\frac{\sqrt{\pi}e^{\alpha^2}}{2}\operatorname{erf}\left(\alpha\right)-\frac{\pi e^{\alpha^2+1}}{4}\left(\operatorname{erf}\left(\beta-\alpha\right)\ \operatorname{erfc}\left(\alpha+\beta\right)\right)+\frac{\pi}{4}}$$
Therefore, the original integral is equal to: $$I=\exp\left(-\frac{a^2+b^2}{8T}\right)f\left(\frac{a}{2\sqrt{2T}},\frac{b}{2\sqrt{2T}}\right)=$$
$$ \bbox[5px,border:2px solid red] { \frac{\pi}{4}\exp\left(-\frac{a^2+b^2}{8T}\right)-\frac{\sqrt{\pi}}{2} \exp\left(-\frac{b^2}{8T}\right)\ \operatorname{erf}\left(\frac{a}{2\sqrt{2T}}\right)-\frac{\pi}{2} \exp\left(-\frac{b^2}{8T}+1\right)\ \operatorname{erf}\left(\frac{b-a}{2\sqrt{2T}}\right)\ \operatorname{erfc}\left(\frac{b+a}{2\sqrt{2T}}\right) } $$