A Help with simplifying an integral

[TroyElliott]

TL;DR

Stuck on some algebra within an integral.

I am not seeing how the v goes away in the third equal sign of equation (1.8). It seems to be that it must be cos(z*sinh(u+v)), not cos(z*sinh(u)).
In the defined equations (1.7), the variable "v" can become imaginary, so a simple change of variables would change the integration sign by adding an imaginary term - not what occurs in (1.8). Any idea on how to get the third line of (1.8)?

Thanks for any insight!

 

[MathematicalPhysicist]

In order to get the third line in Eq. (1.8) change variables $u+v\mapsto u$ and $u-v\mapsto u$.
Since $v$ is constant with respect to the integration varible $du = d(u\pm v)$, and also don't forget that integral of an even function $f(x)=f(-x)$ leads to $\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx$.

 

[TroyElliott]

In order to get the third line in Eq. (1.8) change variables $u+v\mapsto u$ and $u-v\mapsto u$.
Since $v$ is constant with respect to the integration varible $du = d(u\pm v)$, and also don't forget that integral of an even function $f(x)=f(-x)$ leads to $\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx$.

But, in this case, doesn’t it matter than $v$ can potentially become an imaginary number, and the limits of the integral would no longer be just infinity? We would get an added imaginary term to the limit, e.g. $\infty + 2\pi i$.

More explicitly, $v= arcsinh(\frac{mc^{2}t}{hz}),$ where $z$ is given in Eq.(1.7). We see that inside the light cone, $z$ becomes imaginary. So $v$ would also become imaginary, in this case. When performing the change of variables, the integration bound would now pick up an additional imaginary piece. This is what I am confused about - how can it still just be $\infty$?

 

[anuttarasammyak]

u and v are both real and v is constant as for integration of u.
\int_{-\infty}^{+\infty} du = \int_{-\infty}^{+\infty} d(u+v)
or so.

 

[TroyElliott]

In order to get the third line in Eq. (1.8) change variables $u+v\mapsto u$ and $u-v\mapsto u$.
Since $v$ is constant with respect to the integration varible $du = d(u\pm v)$, and also don't forget that integral of an even function $f(x)=f(-x)$ leads to $\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx$.

u and v are both real and v is constant as for integration of u.
\int_{-\infty}^{+\infty} du = \int_{-\infty}^{+\infty} d(u+v)
or so.

But we see in Eq.(1.7), that inside the light cone, $v$ can become imaginary.

 

[anuttarasammyak]

It seems that (1.8) is for real z with u and v real, and (1.9) is for imaginary z.

 

[MathematicalPhysicist]

But we see in Eq.(1.7), that inside the light cone, $v$ can become imaginary.

Such a calculation is done in a course in complex analysis; I don't recall the details, sorry.