How Does a Definite Integral Yield an Imaginary Number?

[paulfr]

Why and how does this definite integral result in an imaginary solution ?

At wolframalpha ...
definite integral 1 / [e^x arcsin x] dx from 1 to 10 = 0.156 + .09i

Area under such a function should be positive or negative but
how does it become imaginary ?

Thanks

 

[awkward]

By usual definition, arcsin(x) is not defined for x > 1.

 

[Feldoh]

By usual definition, arcsin(x) is not defined for x > 1.

It should be clarified that arcsin(x) is not defined for |x| > 1 if x is a real number. Typically, it is defined everywhere on the imaginary axis except for the branch cuts (which are usually taken to be from (-inf,1]U[1,inf).

 

[jackmell]

Why and how does this definite integral result in an imaginary solution ?

At wolframalpha ...
definite integral 1 / [e^x arcsin x] dx from 1 to 10 = 0.156 + .09i

Area under such a function should be positive or negative but
how does it become imaginary ?

Thanks

How:

$$\begin{aligned}<br /> \frac{1}{e^z \arcsin(z)}&=\frac{1}{e^z\left(-i\log[iz+\sqrt{1-z^2}]\right)}\\<br /> &=\frac{1}{e^z\left(-i\log[iz+i\sqrt{z^2-1}]\right)},\quad z>1\\<br /> &=\frac{1}{e^z\left(-i\log[i(z+\sqrt{z^2-1})]\right)}\\<br /> &=\frac{1}{e^z\left(-i(\ln(z+\sqrt{z^2-1})+\pi/2 i)\right)}\\<br /> &=\frac{1}{e^z(\pi/2-i\ln(z+\sqrt{z^2-1}))}\\<br /> &=\frac{\pi/2 e^z+i\ln(z+\sqrt{z^2-1})}{(\pi/2 e^z)^2+e^{2z}\ln^2(z+\sqrt{z^2-1})}<br /> \end{aligned}$$

Now recall:

$$\int_c f(z)dz=\int udx-vdy+i\int udy+vdx$$

and let's just look at the real part of that and since we're on the real-axis during the integration from 1 to 10, then that the real part of that integral is:

$$\int_1^{10} \frac{\pi/2 e^x}{(\pi/2 e^x)^2+e^{2x}\ln^2(x+\sqrt{x^2-1})}\approx 0.158$$

The imaginary part follows similarly. Mathematica as well as the steps above are using the principal log.

 

[paulfr]

Thank you very much

But why does the first line show that

$$arcsin z = \ {\left(-i\log[iz+\sqrt{1-z^2}]\right)}\\$$Where does that come from ?

I appreciate your help, thanks

 

[jackmell]

That comes from the inverse of the complex sine function:

$$w=\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$

If you now solve for z in terms of w, you get that expression, but just use the variable "z" instead of w.