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The function is continuous so the integral exists, but how do you find it :)?
Go ahead and try. Be prepared to be frustrated, however because ...Hey!
Why can't you use the method of intergation by Parts for [tex]\int sin(e^x) dt[/tex]?
This latter integral is encountered in math and science quite frequently, so frequently that it has been given a name -- the sine integral. For more info, see[tex]\int sin(e^t) dt \equiv \int \frac{sinx}{x}dx[/tex]
and that doesn't exist in terms of elementary functions
Why would you think so? If it were exsin(x), f and g would be obvious but here the only functions "multiplied" together are 1 and sin(ex). If you take u= 1 and dv= sin(ex)dx, you are back to the original problem. If you take u= sin(ex) and dv= dx you get du= cos(ex)exdx and v= x so you have gone from [itex]\int udv= \int sin(e^x)dx[/itex] to [itex]uv- \int v du= xsin(e^x)- \int x cos(e^x)e^x dx[/itex] which doesn't look any easier to me.Hey!
Why can't you use the method of intergation by Parts for [tex]\int sin(e^x) dt[/tex]?
We CAN think that this integral is in the form [tex]\int f(x) g(x) dx[/tex], right?
The integral exists, but we can't express it in terms of elementary functions.... that doesn't exist in terms of elementary functions