Step 1: Identify the species reacting.
The question takes sulphur trioxide $SO_3$ and lets it be absorbed into concentrated sulphuric acid $H_2SO_4$. We must name the product.
Step 2: Recall the Contact Process.
In the industrial Contact Process, $SO_3$ is not passed directly into water because the reaction is too violent. Instead it is dissolved in concentrated $H_2SO_4$.
Step 3: Write the addition reaction.
The $SO_3$ adds to $H_2SO_4$: \[ SO_3 + H_2SO_4 \rightarrow H_2S_2O_7. \]
Step 4: Check the formula by atom balance.
Combining $SO_3$ and $H_2SO_4$ gives $2$ sulphur atoms, $2$ hydrogen atoms and $7$ oxygen atoms, which matches $H_2S_2O_7$.
Step 5: Name the product.
$H_2S_2O_7$ is pyrosulphuric acid, commonly called oleum or fuming sulphuric acid.
Step 6: Conclude.
Hence the compound formed when $SO_3$ is absorbed in concentrated $H_2SO_4$ is
\[ \boxed{H_2S_2O_7} \]