If I have:
$$\int_{L}^{\infty }e^{\dfrac{-(x-\sigma \sqrt{T})^2}{2}}\,dx$$
let $y = x - \sigma \sqrt{T}$
$$\int_{L - \sigma \sqrt{T}}^{\infty }e^{\dfrac{-y^2}{2}} \, dy$$
Why does the lower bound change in the new integral?
If I have:
$$\int_{L}^{\infty }e^{\dfrac{-(x-\sigma \sqrt{T})^2}{2}}\,dx$$
let $y = x - \sigma \sqrt{T}$
$$\int_{L - \sigma \sqrt{T}}^{\infty }e^{\dfrac{-y^2}{2}} \, dy$$
Why does the lower bound change in the new integral?
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The lower bound was from $x=L$, after the transformation, we get $y|_{x=L}=x-\sigma\sqrt T |_{x=L} = L-\sigma \sqrt T $, so the lower bound goes from $y=L-\sigma\sqrt T$.
Intuitively, you're performing the integration according to a new variable $y$, which is just a translation of $x$, so you need to translate the boundaries appropriately. Of course, the infinite boundary will remain infinity.