A Help with simplifying an integral

TroyElliott
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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!
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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##.
 
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MathematicalPhysicist said:
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##?
 
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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.
 
MathematicalPhysicist said:
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##.
anuttarasammyak said:
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.
 
It seems that (1.8) is for real z with u and v real, and (1.9) is for imaginary z.
 
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TroyElliott said:
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.
 
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