Proof of a (simple) integro-differential formula

[TeacupPig]

I'm working through a paper and trying to prove the following (dear god... )
\frac{d}{dt}\int_0^t e^{\alpha (t-x)}f(x)dx = f(t) + \alpha\int_0^t e^{\alpha (t-x)}f(x)dx
I have tried three different ways, either by direct manipulations (integration by parts maybe?), or by using the definition of the derivative and taking the limit (should work, but I'm not even getting close), and by switching the order of integration-differentiation (not going to work).

If you manage to show it, please give me some intermediate results or a complete solution if it's short...

 

[SammyS]

I'm working through a paper and trying to prove the following (dear god... )
\frac{d}{dt}\int_0^t e^{\alpha (t-x)}f(x)dx = f(t) + \alpha\int_0^t e^{\alpha (t-x)}f(x)dx
I have tried three different ways, either by direct manipulations (integration by parts maybe?), or by using the definition of the derivative and taking the limit (should work, but I'm not even getting close), and by switching the order of integration-differentiation (not going to work).

If you manage to show it, please give me some hints or a complete solution if it's short...

Try the Leibniz integral rule. Link for Wikipedia article.

 

[TeacupPig]

Thank you very much, the problem is solved by straightforwardly applying this general version of the Leibniz formula.
When walking home I had even thought about it for a brief moment, but I have never seen it before in this form with variable limits (which is what I need).