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Ok this problem really has three parts but its the first part that has me hung up. I have to prove that $\int_0^\pi \frac{\sin(xt)}{t}dt $ is defined.

What I want to do is just show that $\frac{\sin(xt)}{t}$ is continuous on that closed interval for a constant x and therefore integrable, I know I can do this if I set $f(x,t)=1$ for $t=0$, but can I just arbitrarily change the function like that? I could also introduce a dummy variable and integrate from $u \to \pi$ then take the limit as $u$ goes to 0 but I was wondering if having a constant $x$ inside the sin would change the way $\frac{\sin(t)}{t}$converges to 1 as $t \to 0$ .

The second parts I strangely don't find as confusing; find $f'(x)$ ( use leibniz's rule) and show $f'$ is continuous at 0 (can't do this until I've evaluated the integral).

Any help is appreciated, Thanks.

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You dont need to set F(x,t) = 1 at t = 0, look at the limit of function when t tends to 0 – Ram Jan 24 '13 at 6:09

Set it equal to $x$ at $t=0$, not to $1$, since $$\lim_{t\to 0}\frac{\sin xt}{t}=x,$$ which will give you the continuity you want.

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For the second part, use the change of variables $ y=xt $ $$f(x) = \int_0^\pi \frac{\sin(xt)}{t}dt= \int_{0}^{\pi x} \frac{\sin(y)}{y}dt. $$

Now, you can use the Leibniz integral rule.

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