There is one case which is well-known.
Lemma. Let $Q_i \sim \chi^2_{k_i}(\lambda_i)$ for $i=1,\dots,n$, be independent. Then, $Q = \sum_{i=1}^n Q_i$ is a noncentral-$\chi^2_{k}(\lambda)$, where $k=\sum_{i=1}^n k_i$ and $\lambda = \sum_{i=1}^n \lambda_i$.
Proof. The Moment Generating Function (MGF) of $Q_i$ is given by
$$
\mathcal{M}_{Q_i}(t) = (1 -2t)^{-k_i/2} \exp \Big\{ \frac{\lambda_i t}{1 - 2t} \Big\}
$$
The MGF of $Q$, using independence, is given by
$$
\mathcal{M}_Q(t) = \prod_{i=1}^n \mathcal{M}_{Q_i}(t) = \prod_{i=1}^n (1 -2t)^{-k_i/2} \exp \Big\{ \frac{\lambda_i t}{1 - 2t} \Big\} = (1 - 2t)^{- \frac{1}{2} \sum_{i=1}^n k_i} \exp \Big\{ \frac{t}{1 - 2t} \sum_{i=1}^n \lambda_i \Big\}
$$
Thus,
$$
\mathcal{M}_Q(t) = (1 -2t)^{-k/2} \exp \Big\{ \frac{\lambda t}{1 - 2t} \Big\}
$$
and the thesis immediately follows. $\blacksquare$
Note that, if $\lambda \neq 0$, the PDF (probability density function) of $Q$ does not admit a closed-form, as it involves Bessel functions.
The general case of a linear combination of independent $\chi^2_{k_i}(\lambda_i)$
$$
Q = \sum_{i=1}^n a_i Q_i
$$
results in a so-called generalized chi-squared distribution. The probability density function (PDF) of the generalized $\chi^2$ is also complicated.
I have exact the exact motivation of finding the pdf of the chi squared through the characteristic function and got stuck in the same place. But finally I worked it out.
What we want to find is:
$$\int_{-\infty}^{\infty} f(\omega) d\omega=\int_{-\infty}^{\infty} (1 - i2 \omega)^{-k/2} e^{-iwx} d\omega =?$$
For solving the integral, we will use the residue theorem.
The integrand has one pole at -i/2. For calculating the residue of this pole, we will use the Cauchy integral formula for order n and extend it using the concepts of fractional calculus.
The Cauchy integral formula goes as:
$$g^{(n)}(a)=\frac{n!}{2\pi i} \oint_{\gamma} \frac{g(z)}{(z-a)^{n+1}}dz$$
Where $\gamma$ is a closed curve enclosing $a$ and oriented counterclockwise.
Following the principles of fractional calculus, we substitute $n! = \Gamma(n+1)$, rearranging terms and changing the variable n for s to make explicit that we are now in a continuous formula, we have:
$$\oint_{\gamma} \frac{g(z)}{(z-a)^{s}}dz=\frac{2\pi i}{\Gamma(s)}g^{(s-1)}(a)\:,\: s\: \epsilon \: \mathbb{R} $$
In our case, we have $g(z)=e^{-ixz}$, $s=k/2$ and $a=-i/2$.
From the fractional calculus we know that the fractional derivative of the exponential is:
$$\frac{d^{s}}{dz^{s}}e^{az}=a^se^{az}\: \: \Rightarrow \: \: g^{(k)}(z)=(-ix)^ke^{-ixz}$$
Making the substitutions, we get:
$$\oint_{\gamma}\frac{e^{-ixz}}{(1-2iz)^{\frac{k}{2}}}dz=(-2i)^{-\frac{k}{2}}\oint_{\gamma}\frac{e^{-ixz}}{(z+\frac{i}{2})^{\frac{k}{2}}}dz=(-2i)^{-\frac{k}{2}}\frac{2\pi i}{\Gamma(k/2)}(-ix)^{\frac{k}{2}-1}e^{-ix(-i/2)}$$
Simplifying:
$$\oint_{\gamma}\frac{e^{-ixz}}{(1-2iz)^{k/2}}dz=-\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})} = Res(f(z),-\frac{i}{2}) =R$$
Now, we take the integral contour $\gamma$ as in the picture below.
(Note, it seems that I'm not allowed to embed images because I don't have enough reputation)
Integral Contour
[A segment from $r$ to $-r$, with $r\, \epsilon\, \mathbb{R} > 1/2 $, closed by a semicircle centered at the origin with raidus r extending from $\pi$ to $2\pi$. Closed curve oriented counterclockwise. The semicircle path is named $C_{\gamma}^+$, with the + sign indicating the orientation]
If we make $r\rightarrow \infty$, we have:
$$\oint_{\gamma}f(z)dz=\int_{C_{\gamma}^+}f(z)dz-\int_{-\infty}^{\infty}f(\omega)d\omega=R$$
Where the minus in the right hand comes from the integration orientation.
Now, we have to prove that the limit of the integral in $C_{\gamma}^+$ as $r\rightarrow \infty$ is zero:
$$\lim_{r\rightarrow \infty} f(re^{i\theta})=\lim_{r\rightarrow \infty} \frac{e^{-ixre^{i\theta}}}{(1-2ire^{i\theta})^{\frac{k}{2}}} =\lim_{r\rightarrow \infty} \frac{e^{-ixr(\cos(\theta)+i\sin(\theta))}}{(-2i)^{\frac{k}{2}}r^{\frac{k}{2}}e^{\frac{k}{2}i\theta}}$$
$$=(-2i)^{-\frac{k}2{}}\lim_{r\rightarrow \infty}\frac{e^{xr\sin(\theta)}e^{-i(xr\cos(\theta)+\frac{k}{2}\theta)}}{r^{\frac{k}{2}}}=(-2i)^{-\frac{k}2{}}\lim_{r\rightarrow \infty}\frac{e^{xr\sin(\theta)}e^{-ixr\cos(\theta)}}{r^{\frac{k}{2}}}$$
We have used for $r\rightarrow \infty$ $\left | 2ire^{i\theta} \right |\gg 1$ and $\left | xr\cos(\theta) \right |\gg \frac{k}{2}\theta$ for $\theta\neq \pm \pi/2$
The limit of $e^{-ixr\cos(\theta)}$ as $r\rightarrow \infty$ does not exist, however its bounded as the modulus is one, therefore the limit will depend on the other two terms. $r^{k/2}$ tends to infinity, and $e^{xr\sin(\theta)}$ tends to zero for $x>0$
and $\theta \epsilon (\pi,2\pi)$, which is precisely the range corresponding to $C_{\gamma}^+$.
It is immediate to see that for the cases $x=0$ and $\theta=-\pi/2$ the limit is also zero.
Thus we have:
$$\forall \theta \epsilon (\pi,2\pi), x\geq 0 : \lim_{r\rightarrow \infty}f(re^{i\theta})=0$$
$$\int_{C_{\gamma}^+}f(z)dz=0 \Rightarrow \int_{-\infty}^{\infty}f(\omega)d\omega=-R=\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})} $$
Finally, the probability density function of the Chi Squared distribution is:
$$pdf(x,k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-ix\omega}}{(1-i2\omega)^{\frac{k}{2}}}d\omega=\frac{1}{2\pi}\frac{2^{1-\frac{k}{2}}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}\pi}{\Gamma(\frac{k}{2})}=\frac{x^{\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{\frac{k}{2}}\Gamma(\frac{k}2{})}, x\geq0$$
Which is, of course, the value we were looking for.
Best Answer
Let $u=\sqrt{x} \implies \frac{du}{dx}=\frac1{2\sqrt{x}} \implies 2\, du = \frac1{\sqrt{x}}\, dx$. Also $u^2=x$ implies $y-x=y-u^2$. So your integral is
$$\int_0^{\sqrt{y}}\frac{dx}{\sqrt{x(y-x)}}=\int_0^{\sqrt{y}}\frac{2}{\sqrt{y-x}}\cdot \frac1{2\sqrt{x}}\, dx=\int_0^{\sqrt{y}}\frac{2}{\sqrt{y-u^2}}\, du$$ as $0\le x\le \sqrt{y}$. This evaluates to $\left[2\sin^{-1}\left(\frac u{\sqrt{y}}\right)\right]_0^{\sqrt{y}}=2\left(\frac\pi2\right)=\pi$.