Help with difficult (?) integration

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In summary, the conversation involved a problem related to pharmacodynamics and an integral that does not have an analytical solution. The person is not a mathematician and is seeking help to find a solution. They provide a detailed explanation of the problem and equations involved, including the solution for IV administration. However, they are having difficulty finding a solution for PO administration and are seeking suggestions for techniques or methods to find an analytical solution. They also mention trying a change of variables and isolating terms, but are still stuck. They welcome any suggestions for finding a solution or even an approximate solution.
  • #1
lavoisier
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Hi everyone,
I am working on a problem related to pharmacodynamics, and I'm stuck with an integral that doesn't seem to have an analytical solution. Thing is, I'm not a mathematician, so maybe there is a solution and I can't see it. I wonder if anyone can please have a look and tell me.

Here's a summary of the problem (long explanation, sorry - I suppose it can be skipped if not considered relevant).
In pharmacodynamics, it is often found that the percentage biological response (R) depends on the concentration of a drug in plasma (C) according to the equation:

[itex]R(C)=\frac{C^H} {C^H+IC_{50}^H}[/itex]

where IC50 is a constant that measures the strength of the drug-target interaction (the smaller, the stronger the interaction) and H is another constant ('Hill slope') that tells you how fast R varies when C gets close to IC50.
As you can see, when C tends to zero, R also tends to zero; for C=IC50, R is exactly 50%, and when C gets very large, R tends to 100%. This is in agreement with the theory that the biological response is proportional to the percentage of target bound to the drug (under the hypothesis of fast binding kinetics, but that's a different problem).

A parameter that can be derived from R is the integrated efficacy Ieff, which is simply the integral of R over time. For a single dose of drug, the integration is from the time of administration t=0 to infinity.

[itex]I_{eff} = \int_0^{+\inf} R \, dt[/itex]

So one must plug in an expression of R as a function of time (right?).
When you administer a drug, the concentration in plasma vs time is determined, in the simplest cases, by differential equations that express the mass balance of drug entering and exiting the body, and being transformed into something else by metabolic processes.
For intravenous (IV) administration, the solution is:

[itex]C_{IV}(t)=\frac{D_0} {V} e^{- \frac {CL} {V} t}[/itex]

where D0 is the administered dose, V ('volume of distribution') tells you how the drug partitions between plasma and the rest of the body and CL ('clearance') tells you how fast the drug is excreted and metabolised (they are all positive constants).
In this case Ieff can be computed exactly. I got (from Maxima) the following result:

[itex]I_{eff,IV} = \int_0^{+\inf} R(C_{IV}(t)) \, dt = \int_0^{+\inf} \frac{{\big(\frac{D_0} {V} e^{- \frac {CL} {V} t}\big)}^H} {{\big(\frac{D_0} {V} e^{- \frac {CL} {V} t}\big)}^H+IC_{50}^H} \, dt = \frac {V} {H \, CL} Ln \big[ 1+ \big( \frac {D_0} {IC_{50} \, V} \big) ^H \big] [/itex]

which is very useful, because one can then study the effect of the various parameters on Ieff.

For oral (PO) administration, C(t) is more complicated:

[itex]C_{PO}(t)=\frac{D_0 F} {V} \frac{k_a} {k_a-k}(e^{-k t}-e^{-k_a t})[/itex]

The new parameters F and ka are positive constants, F ('bioavailability') expressing how much of the administered dose reaches the systemic circulation, and ka ('first-order absorption constant') expressing how fast the drug is absorbed in the gastro-intestinal tract. k is just CL / V.

I tried to plug this expression into R and integrate, but Maxima couldn't handle it. Indefinite integration gave me something with still an integral inside. Even simplifying to H=1 (most common case) didn't help.

[itex]I_{eff,PO} = \int_0^{+\inf} R(C_{PO}(t)) \, dt [/itex]

So the first question is, do you think the above integral Ieff,PO can be calculated analytically?
And if so, what technique would you suggest?

I would have thought that as C contains exponentials, the ratio R should be reducible to a form like:

[itex]I_{eff} = \int \frac {f'(t)} {f(t)} \, dt[/itex]

(with the addition of appropriate constants of course), but I couldn't find a way to achieve that.
I also considered a series solution, but can one use e.g. Taylor to approximate definite integrals (and especially this one with an infinite limit)?

Thanks!
L
 
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  • #2
lavoisier said:
I am working on a problem related to pharmacodynamics, and I'm stuck with an integral that doesn't seem to have an analytical solution. Thing is, I'm not a mathematician, so maybe there is a solution and I can't see it. I wonder if anyone can please have a look and tell me.
Hi lavoisier:

I noticed in another thread some references that might be helpful.

Regards,
Buzz
 
  • #3
Thank you Buzz Bloom.
I tried Wolfram's integrator, and it didn't find a solution (it also said that it probably means there is no formula for it).

I also tried a change of variable in the special case H=1, in particular:

[itex]y=e^{- k_a t}[/itex]

This gave me a ratio of polynomials in y to integrate, let's call it h(y)/g(y), between the transformed limits 0 and 1, but it didn't help.

I tried to rearrange the ratio in order to isolate the term g'(y)/g(y).
I managed to do that, and I also discovered that the integral of g'(y)/g(y) between 0 and 1 is zero.
However, I was still stuck with a term 1/g(y) that I can't integrate.

[itex]g(y) = a + b (y^c - y) [/itex]

It seems that the term a is the problem, because when a is zero a formula does exist.
I guess if I try another change of variable to make a disappear (assuming that's possible) I will get other non-constant terms in dy.

If anyone can please suggest what else to try, even for an approximate solution...

Thanks
L
 

What is integration and why is it difficult?

Integration is a mathematical process of finding the area under a curve by dividing it into smaller, simpler parts. It is difficult because it requires understanding of various concepts and techniques such as substitution, integration by parts, and trigonometric substitution.

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Practice is key to improving integration skills. Start with simpler integrals and gradually move on to more complex ones. It is also helpful to understand the fundamental concepts and techniques thoroughly.

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