A Derivation of a complex integral with real part

[shardur]

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]

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]

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.

 

[shardur]

Y is real valued.
So my derivative should be correct like I did it?
And thanks for your help!

 

[tnich]

Y is real valued.
So my derivative should be correct like I did it?
And thanks for your help!

It looks right to me. You can simplify it a little if you realize that $exp(i \alpha (b + ln(Y)) - u) = Y^{i \alpha}exp(i \alpha b - u)$

 

[tnich]

It looks right to me. You can simplify it a little if you realize that $exp(i \alpha (b + ln(Y)) - u) = Y^{i \alpha}exp(i \alpha b - u)$

One other thing that is good to know about differentiating integrals is what to do if the variable of differentiation appears in the limits of the integral as well as in the integrand. If you have an integral that looks like this
$$F(y) = \int_{g(y)}^{h(y)}f(y,x)~dx$$
and you want to differentiate with respect to y, then you get
$$\frac {dF(y)} {dy} = f(y,h(y))\frac {dh(y)} {dy}-f(y,g(y))\frac {dg(y)} {dy}+\int_{g(y)}^{h(y)}\frac{df(y,x)}{dy}~dx$$
That is, you need to differentiate with respect to all of the occurrences of y in the integral.