Solving Integral Equations: Find x from 1-x+ ∫^x_1 (sin t/t) dt

Rectifier
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The problem
I want to find ##x## which solves ## 1-x+ \int^x_1 \frac{\sin t}{t} \ dt = 0 ##

The attempt
##\int^x_1 \frac{\sin t}{t} \ dt = x -1 ## I see that the answer is ##x=1## but I want to be able to calculate it mechanically in case if I get similar problem with other elements. Any suggestions on how I can do that?
 
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Hi,
'mechanically' sounds good. But, ##{\sin x\over x}## is (and I https://owlcation.com/stem/How-to-Integrate-sinxx-and-cosxx )

one of the simplest examples of non-integrable functions in the sense that their antiderivatives cannot be expressed in terms of elementary functions, in other words, they don't have closed-form antiderivatives.​

However, apart from ##x=1## there shouldn't be too many other solutions ... ##x-1## grows faster than the integral.
You could also investigate domain [0,x] : with ##{\sin x\over x} < 1## the integral is always different from x-1.
 
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Okay, thank your for your help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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