I Integration being unchanged after rotation

[AndrewGRQTF]

This question is about the general 1 loop correction to the propagator in QFT (this is actually not important for this question). Let's say we have an integral over an integration variable x, and this x ranges from $-\infty$ to $\infty$. If we look at this integration contour in the complex plane, it will be along the real axis. A book that I am reading (QFT by Srednicki), says that we can rotate this contour counterclockwise onto the imaginary axis, without changing the value of the integral, and he says that this is because the contour does not pass over any poles of the integrand and that the integrand vanishes fast enough as the magnitude of x goes to infinity.

My question is: why does the value of the integral not change when we change the integration contour?

 

[mfb]

The integral of an entire function over a closed path is zero if there are no poles inside (residue theorem). If the function goes to zero sufficiently fast for large |z| you can write the original integral as sum of the new (rotated) integral plus two closed paths (two 90 degree sectors of the complex plane https://www.researchgate.net/figure/Integration-contour-in-the-complex-s-plane-to-compute-the-integral-representing-the_fig3_242423238, then let their radius go to infinity), the contribution from the closed paths is zero.

 

[AndrewGRQTF]

The integral of an entire function over a closed path is zero if there are no poles inside (residue theorem). If the function goes to zero sufficiently fast for large |z| you can write the original integral as sum of the new (rotated) integral plus two closed paths (two 90 degree sectors of the complex plane https://www.researchgate.net/figure/Integration-contour-in-the-complex-s-plane-to-compute-the-integral-representing-the_fig3_242423238, then let their radius go to infinity), the contribution from the closed paths is zero.

Thank you.