Show that √01dx/√(-logx) =√¶
Answers
Answer:
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Step-by-step explanation:
How do you prove this? ∫10dx(−lnx)−−−−−−√=π−−√
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The function
f(x)=1ln(−ln(x))−−−−−−−−−√
is real-valued for
0<x<1e.
For x=1e,ln(−ln(x))−−−−−−−−−√=0 and f(x) is indeterminate.
For x>1e,f(x) is negative and complex valued.
The results above are represented in the following plot (made with Mathematica). It can be seen that f(x) has a singularity or singular point at x=1e≈0.3678794411714423215955238:
Taking into account the singularity, the integral in the question can be expressed as:
[math]\displaystyle {\int_[/math]
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Given the other answers, we can be certain that this integral is wrong, but it does seem to be potentially correct. So let’s try to derive the correct equation.
From first glance, since the right hand side is π−−√ , the integral probably originates from the Gaussian integral.
∫∞−∞e−x2dx=π−−√
Looking at this integral, it’s convincing that this must be where it’s derived from: the Gaussian integral has an exponential, the above integral has a log; the Gaussian integral has a square, the above integral has a square root.
For simplicity later down the line, we will use the integral
∫∞0e−x2dx=π√2
Using u-substitution, we set x=−u−−−√ and dx=−12−u√du , and the bounds are u0= 0 and u∞=− ∞2=−∞ the integral changes to
∫−∞0−eu2−u√du=π√2
Renaming u to x
∫−∞0−ex2−x√dx=π√2
Now, u substituting again, we set u=ex , x=lnu , and dx=1udu , and the bounds change to u0=e0=1 and u−∞=e−∞=0 , the integral becomes
∫01−u2−lnu√1udu=π√2
Swapping the top and bottom conditions and simplifying gets
∫1012−lnu√du=π√2
And finally, multiplying both sides by two gives the correct integral
∫10dx−lnx√=π−−√
which is very close to the integral in the question, so the original integral with the extra ln was probably just a typo.