# Derivation of a complex integral with real part

• A
Hey,

I tried to construct the derivation of the integral C with respect to Y:

$$\frac{\partial C}{\partial Y} = ?$$
$$C = \frac{2}{\pi} \int_0^{\infty} Re(d(\alpha) \frac{exp(-i \cdot ln(f))}{i \alpha}) d \alpha$$
with
$$d(\alpha) = exp(i \alpha (b + ln(Y)) - u) \cdot exp(v(\alpha) + z (\alpha))$$
where ##Re()## is the real part of whatever is written in the brackets, ##i## is the imaginary number and ##exp()## is the exponential function of whatever is written in the brackets. ##d(\alpha), v(\alpha), z(\alpha)## are functions depending on ##\alpha##. However, ##v(\alpha), z(\alpha)## are not dependent on ##Y##.
How do I differentiate this complex integral? As only ##d(\alpha)## depends on ##Y##, do I just need to differentiate ##d(\alpha)## with respect to ##Y## and plug the outcome of that into my integral? I think that would be what the Leibniz rule would suggest. So like:
$$\frac{\partial C}{\partial Y} = \frac{2}{\pi} \int_0^{\infty} Re(\frac{\partial d(\alpha)}{\partial Y} \frac{exp(-i \cdot ln(f))}{i \alpha}) d \alpha$$
Would that be correct? Because then
$$\frac{\partial d(\alpha)}{\partial Y} = \frac{i \alpha}{Y} \cdot exp(i \alpha (b + ln(Y)) - u) \cdot exp(v(\alpha) + z (\alpha))$$
Would that be the correct way to differentiate the integral or am I doing it completely wrong?

I hope I defined everything properly, if not just let me know.

Thanks a lot for your help!

tnich
Homework Helper
Hey,

I tried to construct the derivation of the integral C with respect to Y:

$$\frac{\partial C}{\partial Y} = ?$$
$$C = \frac{2}{\pi} \int_0^{\infty} Re(d(\alpha) \frac{exp(-i \cdot ln(f))}{i \alpha}) d \alpha$$
with
$$d(\alpha) = exp(i \alpha (b + ln(Y)) - u) \cdot exp(v(\alpha) + z (\alpha))$$
where ##Re()## is the real part of whatever is written in the brackets, ##i## is the imaginary number and ##exp()## is the exponential function of whatever is written in the brackets. ##d(\alpha), v(\alpha), z(\alpha)## are functions depending on ##\alpha##. However, ##v(\alpha), z(\alpha)## are not dependent on ##Y##.
How do I differentiate this complex integral? As only ##d(\alpha)## depends on ##Y##, do I just need to differentiate ##d(\alpha)## with respect to ##Y## and plug the outcome of that into my integral? I think that would be what the Leibniz rule would suggest. So like:
$$\frac{\partial C}{\partial Y} = \frac{2}{\pi} \int_0^{\infty} Re(\frac{\partial d(\alpha)}{\partial Y} \frac{exp(-i \cdot ln(f))}{i \alpha}) d \alpha$$
Would that be correct? Because then
$$\frac{\partial d(\alpha)}{\partial Y} = \frac{i \alpha}{Y} \cdot exp(i \alpha (b + ln(Y)) - u) \cdot exp(v(\alpha) + z (\alpha))$$
Would that be the correct way to differentiate the integral or am I doing it completely wrong?

I hope I defined everything properly, if not just let me know.

Thanks a lot for your help!
I think that would be correct if Y is real-valued. If it is not, then I am not sure that ##\frac d {dY} Re(g(Y)) = Re(\frac d {dY} g(Y))##.

tnich
Homework Helper
I think that would be correct if Y is real-valued. If it is not, then I am not sure that ##\frac d {dY} Re(g(Y)) = Re(\frac d {dY} g(Y))##.
I seem to be wrong about that. It looks to me like it works for complex values of Y, also.

Y is real valued.
So my derivative should be correct like I did it?

tnich
Homework Helper
Y is real valued.
So my derivative should be correct like I did it?
$$F(y) = \int_{g(y)}^{h(y)}f(y,x)~dx$$
$$\frac {dF(y)} {dy} = f(y,h(y))\frac {dh(y)} {dy}-f(y,g(y))\frac {dg(y)} {dy}+\int_{g(y)}^{h(y)}\frac{df(y,x)}{dy}~dx$$