Notation pedantry (integration by substitution)? In a summative assessment, I lost a mark due to this:
$$f_X(x)=\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty  y^2\exp\left\{-\left(\frac{1}{2}+\lambda x\right)y^2\right\} \; dy$$
Now let $a=\frac{1}{2}+\lambda x$ for brevity and $u=ay^2$ so $du=2ay \; du$
Then
$$f_X(x)=\frac{2\lambda}{\sqrt{2\pi}}\int_0^\infty \frac{u}{a} e^{-u} \; du.$$
The comment was "show that you have considered any possible change in limits", and "y=" and "u=" were inserted into my script, so that it now reads:
$$f_X(x)=\frac{\lambda}{\sqrt{2\pi}}\int_{y=-\infty}^{y=\infty} y^2\exp\left\{-\left(\frac{1}{2}+\lambda x\right)y^2\right\} \; dy$$
Then
$$f_X(x)=\frac{2\lambda}{\sqrt{2\pi}}\int_{u=0}^{u=\infty}\frac{u}{a} e^{-u} \; du.$$
This is not a calculus module (it's statistical theory, final year undergraduate), and I feel pretty hard done by ! Is this overly pedantic on the part of the marker ?
EDIT, rather than correct the typos in this post, which would obscure robjohn's answer, I have posted an answer which (hopefully) doesn't contain any typos and makes the point more clearly. 
 A: This is not pedantic! The marker is trying to point out where you went wrong.
I will assume that your integral is actually
$$
\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y\tag{1}
$$
First of all, if $u=ay^2$ and $y=-\infty$, it would seem that $u=\infty$, not $0$.
Secondly, if you are going to use $u=ay^2$, you should break up the integral into pieces that will have disjoint domains. Note that under $u=ay^2$ both $(-\infty,0]$ and $[0,\infty)$ get mapped to $[0,\infty)$. To prevent confusion, the integral should first be broken up into
$$
\small
\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y
=\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^0 y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y
+\frac{\lambda}{\sqrt{2\pi}}\int_0^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y\tag{2}
$$
One might then notice that since $y^2e^{-(1/2+\lambda x)y^2}$ is even, the two integrals on the right in $(2)$ are equal, therefore
$$
\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y
=\frac{2\lambda}{\sqrt{2\pi}}\int_0^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y\tag{3}
$$
Finally, we can apply $u=ay^2$:
$$
\begin{align}
\frac{2\lambda}{\sqrt{2\pi}}\int_0^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y
&=\frac{2\lambda}{\sqrt{2\pi}}\int_0^\infty\frac{u}{a}e^{-u}\frac{\mathrm{d}u}{2ay}\\
&=\frac{\lambda}{\sqrt{2\pi}}\int_0^\infty\frac{u}{a}e^{-u}\frac{\mathrm{d}u}{\sqrt{au}}\tag{4}
\end{align}
$$
Note the correct translation of $\mathrm{d}y$. Therefore, the substitution should yield
$$
\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty y^2e^{-(1/2+\lambda x)y^2}\mathrm{d}y
=\frac{\lambda}{\sqrt{2\pi}}\int_0^\infty\frac{u}{a}e^{-u}\frac{\mathrm{d}u}{\sqrt{au}}\tag{5}
$$
A: Your marker was not suggesting that you should write the $y=$ and $u=$ signs on the integrals (I like to do that myself, but there's no reason you should as it's perfectly clear from the $d y$ and the $du$ that those are the variables being integrated over).  Rather, (s)he was trying to point out that you should have inserted an extra step noting explicitly that $y=\infty$ corresponds to $u=\infty$ and that $y=-\infty$ corresponds to $u=0$ - considering that this is an undergraduate course, I don't really see why you should have put this in since it seems perfectly clear and you did put the right limits into the integrals, so I would feel a little hard done by as well.
A: FWIW, this is my complete answer to the question:
$$f_X(x)=\frac{\lambda}{\sqrt{2\pi}}\int_{-\infty}^\infty y^2 \text{exp}(-(1/2+\lambda x)y^2)\mathrm{d}y$$
Let $a=\frac{1}{2}+\lambda x$ for brevity and $u=ay^2$ so $du=2ay \mathrm{ d}y$. Then
$$f_X(x)=\frac{2\lambda}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{u}{a} e^{-u} \frac{\mathrm{d}u}{2\sqrt{au}}$$
using the symmetry of the integrand. So
$$
\begin{align}
f_X(x) 

&= \frac{\lambda a^{-\frac{3}{2}}}{\sqrt{2\pi}}\int_{0}^{\infty}u^{\frac{3}{2}-1}e^u \mathrm{d}u\\
&= \frac{\lambda a^{-\frac{3}{2}}\Gamma(3/2)}{\sqrt{2\pi}}\\
&= \frac{\lambda}{2 \sqrt{2}(\frac{1}{2}+\lambda x)^{\frac{3}{2}}}
\end{align}
$$
I got 6 out of 7 marks - with 1 mark deducted for not writing y= and u= in the integration limits, and this is what my OP was about - I'm sorry for the typos in the OP ! 
