How do you know where to place the new parameter in the original integrand to use feynman's integral trick?

[Al-Layth]

In feynman's integral trick which involves differentiating under the integral sign, we need to first add another parameter to define a new function I(t). here are some example problems of it being done.

[math]\int_{0}^{1} \ln(x)dx \rightarrow I(t)=\int_{0}^{1} \ln(tx)dx[/math]
[math]\int_{0}^{ \infty} \frac{\sin(x)}{x}dx \longrightarrow I(t)=\int_{0}^{ \infty} \frac{\sin(tx)}{x}dx[/math]
I don't understand why they placed the parameter where they did, why not elsewhere, and what reasoning a person uses to decide where to place the parameter.

Can someone help me understand
thanks

 

[topsquark]

In feynman's integral trick which involves differentiating under the integral sign, we need to first add another parameter to define a new function I(t). here are some example problems of it being done.

[math]\int_{0}^{1} \ln(x)dx \rightarrow I(t)=\int_{0}^{1} \ln(tx)dx[/math]
[math]\int_{0}^{ \infty} \frac{\sin(x)}{x}dx \longrightarrow I(t)=\int_{0}^{ \infty} \frac{\sin(tx)}{x}dx[/math]
I don't understand why they placed the parameter where they did, why not elsewhere, and what reasoning a person uses to decide where to place the parameter.

Can someone help me understand
thanks

There's a bit of experience where you put the parameter but generally it goes inside a "function" ie. inside sin(x) or ln(x) or [imath]e^x[/imath] rather than next to an x. In the long run the best way to get some experience with it is to try it out on a few integrals. It's somewhat time consuming, but educational!

-Dan

 

[Al-Layth]

There's a bit of experience where you put the parameter but generally it goes inside a "function" ie. inside sin(x) or ln(x) or [imath]e^x[/imath] rather than next to an x. In the long run the best way to get some experience with it is to try it out on a few integrals. It's somewhat time consuming, but educational!

-Dan

Okay
Thank you