Solve the integral equation of convolution (Abel) 
Solve the integral equation, $$I = \int_{0}^{t} \frac{f(\tau)}{\sqrt{t - \tau}}d\tau = \sqrt{2g}T$$ where $T, g$ are constants. Find $f(t)$

I see that $$I = f(t) * g(t)$$ where $f(t)$ we need to find and $g(t) = \frac{1}{\sqrt{t}}$
Using the convolution theorem, we see that:
$$\sqrt{2g}T = \mathcal{L}^{-1}\{ \mathcal{L}(f) \mathcal{L}(g) \} $$
Let $F(s) = \mathcal{L}(f)$ and we know $\mathcal{L}(g) = \frac{\sqrt{\pi}}{2s^{3/2}}$ therefore,
$$\sqrt{2g}T = \frac{\sqrt{\pi}}{2}\mathcal{L}^{-1} \{\frac{F(s)}{s^{3/2}}\}$$
Now, I got stuck.
Can someone provide some help?
 A: Start by taking the Laplace transform of both sides:
$$\mathcal{L}\left\{f(t)\ast \frac{1}{\sqrt{t}}\right\}=\mathcal{L}\left\{\sqrt{2g}\cdot T\right\} \tag{1}$$
Note that the Laplace transform of $\dfrac{1}{\sqrt{t}}$ is not $\dfrac{\sqrt{\pi}}{2s^{3/2}}$, that is the Laplace transform of $\sqrt{t}$. Instead, you should have the following, since $\mathcal{L}\left\{\dfrac{1}{\sqrt{t}}\right\}=\dfrac{\sqrt{\pi}}{\sqrt{s}}$.
$$F(s)\cdot \color{green}{\frac{\sqrt{\pi}}{\sqrt{s}}}=\frac{\sqrt{2g}\cdot T}{s}$$
Where $F(s)=\mathcal{L}\{f(t)\}$. Now, it remains to solve for $F(s)$:
$$F(s)=\frac{\sqrt{2g}\cdot T}{\sqrt{\pi}}\cdot \frac{1}{\sqrt{s}} \tag{2}$$
Evaluate the Inverse Laplace transform of that to obtain $f(t)$. If you would like to verify your answer, check if it satisfies the integral equation: I tried it, it works.
A: As you did, using the convolution theorem:
$$\mathscr{L}_t\left[\text{f}\left(t\right)\space*\space\text{g}\left(t\right)\right]_{\left(\text{s}\right)}=\text{F}\left(\text{s}\right)\cdot\text{G}\left(\text{s}\right)\tag1$$
So, we get:
$$\mathscr{L}_t\left[\int_0^t\frac{\text{f}\left(\tau\right)}{\sqrt{t-\tau}}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\mathscr{L}_t\left[\text{f}\left(t\right)\space*\space\frac{1}{\sqrt{t}}\right]_{\left(\text{s}\right)}=\text{F}\left(\text{s}\right)\cdot\frac{\sqrt{\pi}}{\sqrt{\text{s}}}=$$
$$\mathscr{L}_t\left[\text{T}\cdot\sqrt{2\text{g}}\right]_{\left(\text{s}\right)}=\text{T}\cdot\sqrt{2\text{g}}\cdot\mathscr{L}_t\left[1\right]_{\left(\text{s}\right)}=\frac{\text{T}\cdot\sqrt{2\text{g}}}{\text{s}}\tag2$$
So, we get:
$$\text{F}\left(\text{s}\right)\cdot\frac{\sqrt{\pi}}{\sqrt{\text{s}}}=\frac{\text{T}\cdot\sqrt{2\text{g}}}{\text{s}}\space\Longleftrightarrow\space\text{F}\left(\text{s}\right)=\frac{1}{\sqrt{\text{s}}}\cdot\frac{\text{T}\cdot\sqrt{2\text{g}}}{\sqrt{\pi}}\tag3$$
And the inverse Laplace transform:
$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\sqrt{\text{s}}}\right]_{\left(t\right)}=\frac{1}{\sqrt{\pi}\cdot\sqrt{t}}\tag4$$
So, we get:
$$\text{f}\left(t\right)=\frac{1}{\sqrt{\pi}\cdot\sqrt{t}}\cdot\frac{\text{T}\cdot\sqrt{2\text{g}}}{\sqrt{\pi}}=\frac{\text{T}\cdot\sqrt{2\text{g}}}{\pi}\cdot\frac{1}{\sqrt{t}}\tag5$$
