Log (sinx) von 0 nach pi / 2 einbinden?

Antworten:

$I = \int_{0}^{\frac{π}{2}} log sin x d x = - (\frac{π}{2}) log 2$ $I=int_0^(pi/2)logsinxdx=-(pi/2)log2$

Erläuterung:

Wir nutzen das Grundstück $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ $int_0^af(x)dx=int_0^af(a-x)dx$

daher können wir schreiben $I = \int_{0}^{\frac{π}{2}} log sin x d x = \int_{0}^{\frac{π}{2}} log sin (\frac{π}{2} - x) d x$ $I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logsin(pi/2-x)dx$

or $I = \int_{0}^{\frac{π}{2}} log sin x d x = \int_{0}^{\frac{π}{2}} log cos x d x$ $I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logcosxdx$

or $2 I = \int_{0}^{\frac{π}{2}} (log sin x + log cos x) d x = \int_{0}^{\frac{π}{2}} log (sin x cos x) d x$ $2I=int_0^(pi/2)(logsinx+logcosx)dx=int_0^(pi/2)log(sinxcosx)dx$

= $\int_{0}^{\frac{π}{2}} log (\frac{sin 2 x}{2}) d x = \int_{0}^{\frac{π}{2}} (log sin 2 x - log 2) d x$ $int_0^(pi/2)log((sin2x)/2)dx=int_0^(pi/2)(logsin2x-log2)dx$

= $\int_{0}^{\frac{π}{2}} log sin 2 x d x - \int_{0}^{\frac{π}{2}} log 2 d x$ $int_0^(pi/2)logsin2xdx-int_0^(pi/2)log2dx$

= $\int_{0}^{\frac{π}{2}} log sin 2 x d x - (\frac{π}{2}) log 2$ $int_0^(pi/2)logsin2xdx-(pi/2)log2$ .............(EIN)

Lassen $I_{1} = \int_{0}^{\frac{π}{2}} log sin 2 x d x$ $I_1=int_0^(pi/2)logsin2xdx$ und $t = 2 x$ $t=2x$ , dann $I_{1} = \frac{1}{2} \int_{0}^{π} log sin t d t$ $I_1=1/2int_0^pilogsintdt$

und mit der Eigenschaft $\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (a - x) d x$ $int_0^(2a)f(x)dx=2int_0^af(a-x)dx$ Wenn $f (2 a - x) = f (x)$ $f(2a-x)=f(x)$ - Beachten Sie das hier $log sin t = log sin (π - t)$ $logsint=logsin(pi-t)$ und wir bekommen

$I_{1} = \frac{1}{2} \int_{0}^{π} log sin t d t = \int_{0}^{\frac{π}{2}} log sin t d t = I$ $I_1=1/2int_0^pilogsintdt=int_0^(pi/2)logsintdt=I$

Daher (A) wird $2 I = I - (\frac{π}{2}) log 2$ $2I=I-(pi/2)log2$

or $I = - (\frac{π}{2}) log 2$ $I=-(pi/2)log2$