Question

please solve it.........​

Answer 1

Question :

Prove :

$\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}=\sqrt{\pi}$

Solution :

We have ,

$\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}$

Let $\sf\:u=-\log\:x$

Now Differentiate with respect to x

$\sf\dfrac{du}{dx}=\dfrac{-1}{x}$

$\sf\:dx=-xdu$

Also ,

$\sf\:u=-\log\:x$

$\sf\implies\:e^{-u}=x$

• When x = 0

⇒u=∞

•When x= 1

⇒u=0

Now ,

$\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}$

$\sf=\int\limits_{\infty}^0\dfrac{1}{\sqrt{u}}\times-xdu$

$\sf=\int\limits_{\infty}^0\dfrac{-e^{-u}}{\sqrt{u}}du$

Now let $\sf\sqrt{u}=v$

Now Differentiate with respect to x

$\sf\:du=2\sqrt{u}dv$

Also,

$\sf\sqrt{u}=v$

• When x = 0

⇒u=∞

•When x= 1

⇒u=0

Now ,

$\sf\int\limits_{\infty}^0\dfrac{-e^{-u}}{\sqrt{u}}du$

$\sf=\int\limits_{\infty}^0\dfrac{-e^{-v^2}}{v}\times2\sqrt{u}dv$

$\sf=\int\limits_{\infty}^0\:-2e^{-v^2}dv$

Now multiply and divide by √π

$\sf=\dfrac{\sqrt{\pi}}{\sqrt{\pi}}\int\limits_{\infty}^0-2e^{-v^2}dv$

$\sf=\sqrt{\pi}\times-\int\limits_{\infty}^0\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv$

We know that

$\sf\int\limits_{a}^bf(x)dx=-\int\limits_{b}^af(x)dx$ i.e ,the interchange of limits of a definite integral changes only its sign. Then,

$\sf=\sqrt{\pi}\int\limits_{0}^{\infty}\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv$

Now the integral changes in special(non-elementary ) type of integral ,which is known as Error Function

Error Function defined as :

$\sf\:erf(x)=\dfrac{2}{\sqrt{\pi}}\int\limits_{0}^x\:e^{-t^2}dt$

Thus,

$\sf\sqrt{\pi}\int\limits_{0}^{\infty}\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv$

$\sf=\sqrt{\pi}\:erf(\infty)$

We know that erf(∞)=1

$\sf=\sqrt{\pi}$

Hence, Proved

$\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}=\sqrt{\pi}$

$\rule{200}2$

More About the topic :

${\purple{\boxed{\large{\bold{Error\: Function}}}}}$

The error function (also called the Gauss error function), often denoted by erf,

It is defined as:

$\sf\:erf(x)=\dfrac{2}{\sqrt{\pi}}\int\limits_{0}^x\:e^{-t^2}dt$

Properties :

1) erf(∞)=1

2) Domain (-∞,∞)

Answer 2

Answer:

Question :

Prove :

\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}=\sqrt{\pi}

0

∫

1

−logx

dx

=

π

Solution :

We have ,

\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}

0

∫

1

−logx

dx

Let \sf\:u=-\log\:xu=−logx

Now Differentiate with respect to x

\sf\dfrac{du}{dx}=\dfrac{-1}{x}

dx

du

=

x

−1

\sf\:dx=-xdudx=−xdu

Also ,

\sf\:u=-\log\:xu=−logx

\sf\implies\:e^{-u}=x⟹e

−u

=x

• When x = 0

⇒u=∞

•When x= 1

⇒u=0

Now ,

\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}

0

∫

1

−logx

dx

\sf=\int\limits_{\infty}^0\dfrac{1}{\sqrt{u}}\times-xdu=

∞

∫

0

u

1

×−xdu

\sf=\int\limits_{\infty}^0\dfrac{-e^{-u}}{\sqrt{u}}du=

∞

∫

0

u

−e

−u

du

Now let \sf\sqrt{u}=v

u

=v

Now Differentiate with respect to x

\sf\:du=2\sqrt{u}dvdu=2

u

dv

Also,

\sf\sqrt{u}=v

u

=v

• When x = 0

⇒u=∞

•When x= 1

⇒u=0

Now ,

\sf\int\limits_{\infty}^0\dfrac{-e^{-u}}{\sqrt{u}}du

∞

∫

0

u

−e

−u

du

\sf=\int\limits_{\infty}^0\dfrac{-e^{-v^2}}{v}\times2\sqrt{u}dv=

∞

∫

0

v

−e

−v

2

×2

u

dv

\sf=\int\limits_{\infty}^0\:-2e^{-v^2}dv=

∞

∫

0

−2e

−v

2

dv

Now multiply and divide by √π

\sf=\dfrac{\sqrt{\pi}}{\sqrt{\pi}}\int\limits_{\infty}^0-2e^{-v^2}dv=

π

π

∞

∫

0

−2e

−v

2

dv

\sf=\sqrt{\pi}\times-\int\limits_{\infty}^0\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv=

π

×−

∞

∫

0

π

2e

−v

2

dv

We know that

\sf\int\limits_{a}^bf(x)dx=-\int\limits_{b}^af(x)dx

a

∫

b

f(x)dx=−

b

∫

a

f(x)dx i.e ,the interchange of limits of a definite integral changes only its sign. Then,

\sf=\sqrt{\pi}\int\limits_{0}^{\infty}\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv=

π

0

∫

∞

π

2e

−v

2

dv

Now the integral changes in special(non-elementary ) type of integral ,which is known as Error Function

Error Function defined as :

\sf\:erf(x)=\dfrac{2}{\sqrt{\pi}}\int\limits_{0}^x\:e^{-t^2}dterf(x)=

π

2

0

∫

x

e

−t

2

dt

Thus,

\sf\sqrt{\pi}\int\limits_{0}^{\infty}\dfrac{2e^{-v^2}}{\sqrt{\pi}}dv

π

0

∫

∞

π

2e

−v

2

dv

\sf=\sqrt{\pi}\:erf(\infty)=

π

erf(∞)

We know that erf(∞)=1

\sf=\sqrt{\pi}=

π

Hence, Proved

\sf\int\limits_{0}^1\dfrac{dx}{\sqrt{-\log\:x}}=\sqrt{\pi}

0

∫

1

−logx

dx

=

π

\rule{200}2

More About the topic :

{\purple{\boxed{\large{\bold{Error\: Function}}}}}

ErrorFunction

The error function (also called the Gauss error function), often denoted by erf,

It is defined as:

\sf\:erf(x)=\dfrac{2}{\sqrt{\pi}}\int\limits_{0}^x\:e^{-t^2}dterf(x)=

π

2

0

∫

x

e

−t

2

dt

Properties :

1) erf(∞)=1

2) Domain (-∞,∞)

Step-by-step explanation:

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