Does $(alpha=int_{-infty}^{infty} f(x), mathrm{d}x)$ follow from $(lim_{kto infty} int_{a_k}^{b_k}...
Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:
When an $(alphain mathbb{R})$ exists, so that for all sequences $((a_k)_{kin mathbb{N}}, (b_k)_{kin mathbb{N}} subsetmathbb{R})$ with $(a_kto -infty, b_kto infty)$ for $(kto infty)$ already $(lim_{kto infty} int_{a_k}^{b_k} f(x),mathrm{d}x=alpha)$ applies, then the following $(alpha=int_{-infty}^{infty} f(x), mathrm{d}x)$ applies too.
My thoughts are that when $(a_kto -infty, b_kto infty)$ for $(kto infty)$ applies, then at some time (meaning for $(kgeq k_0)$ with a $(k_0in mathbb{N}))$ the following: $(a_k<0)$ and $(b_k>0)$ hold.
So the integral can be split like that: $(int_{a_k}^{b_k} f(x)mathrm{d}x=int_{a_k}^{0} f(x)mathrm{d}x+int_{0}^{b_k} f(x)mathrm{d}x)$ which leads us to the definition of improper integrals.
Is my argumentation correct or wrong? Because from $(lim_{nto infty} (c_n+d_n))$ the existence of $(lim_{nto infty}c_n)$ or $(lim_{nto infty}d_n))$ do not follow...
real-analysis integration limits improper-integrals
edited Dec 4 '18 at 23:51
MOMO
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
add a comment |
Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:
When an $(alphain mathbb{R})$ exists, so that for all sequences $((a_k)_{kin mathbb{N}}, (b_k)_{kin mathbb{N}} subsetmathbb{R})$ with $(a_kto -infty, b_kto infty)$ for $(kto infty)$ already $(lim_{kto infty} int_{a_k}^{b_k} f(x),mathrm{d}x=alpha)$ applies, then the following $(alpha=int_{-infty}^{infty} f(x), mathrm{d}x)$ applies too.
My thoughts are that when $(a_kto -infty, b_kto infty)$ for $(kto infty)$ applies, then at some time (meaning for $(kgeq k_0)$ with a $(k_0in mathbb{N}))$ the following: $(a_k<0)$ and $(b_k>0)$ hold.
So the integral can be split like that: $(int_{a_k}^{b_k} f(x)mathrm{d}x=int_{a_k}^{0} f(x)mathrm{d}x+int_{0}^{b_k} f(x)mathrm{d}x)$ which leads us to the definition of improper integrals.
Is my argumentation correct or wrong? Because from $(lim_{nto infty} (c_n+d_n))$ the existence of $(lim_{nto infty}c_n)$ or $(lim_{nto infty}d_n))$ do not follow...
real-analysis integration limits improper-integrals
edited Dec 4 '18 at 23:51
MOMO
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
add a comment |
Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:
When an $(alphain mathbb{R})$ exists, so that for all sequences $((a_k)_{kin mathbb{N}}, (b_k)_{kin mathbb{N}} subsetmathbb{R})$ with $(a_kto -infty, b_kto infty)$ for $(kto infty)$ already $(lim_{kto infty} int_{a_k}^{b_k} f(x),mathrm{d}x=alpha)$ applies, then the following $(alpha=int_{-infty}^{infty} f(x), mathrm{d}x)$ applies too.
My thoughts are that when $(a_kto -infty, b_kto infty)$ for $(kto infty)$ applies, then at some time (meaning for $(kgeq k_0)$ with a $(k_0in mathbb{N}))$ the following: $(a_k<0)$ and $(b_k>0)$ hold.
So the integral can be split like that: $(int_{a_k}^{b_k} f(x)mathrm{d}x=int_{a_k}^{0} f(x)mathrm{d}x+int_{0}^{b_k} f(x)mathrm{d}x)$ which leads us to the definition of improper integrals.
Is my argumentation correct or wrong? Because from $(lim_{nto infty} (c_n+d_n))$ the existence of $(lim_{nto infty}c_n)$ or $(lim_{nto infty}d_n))$ do not follow...
real-analysis integration limits improper-integrals
edited Dec 4 '18 at 23:51
MOMO
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:
When an $(alphain mathbb{R})$ exists, so that for all sequences $((a_k)_{kin mathbb{N}}, (b_k)_{kin mathbb{N}} subsetmathbb{R})$ with $(a_kto -infty, b_kto infty)$ for $(kto infty)$ already $(lim_{kto infty} int_{a_k}^{b_k} f(x),mathrm{d}x=alpha)$ applies, then the following $(alpha=int_{-infty}^{infty} f(x), mathrm{d}x)$ applies too.
My thoughts are that when $(a_kto -infty, b_kto infty)$ for $(kto infty)$ applies, then at some time (meaning for $(kgeq k_0)$ with a $(k_0in mathbb{N}))$ the following: $(a_k<0)$ and $(b_k>0)$ hold.
So the integral can be split like that: $(int_{a_k}^{b_k} f(x)mathrm{d}x=int_{a_k}^{0} f(x)mathrm{d}x+int_{0}^{b_k} f(x)mathrm{d}x)$ which leads us to the definition of improper integrals.
Is my argumentation correct or wrong? Because from $(lim_{nto infty} (c_n+d_n))$ the existence of $(lim_{nto infty}c_n)$ or $(lim_{nto infty}d_n))$ do not follow...
real-analysis integration limits improper-integrals
real-analysis integration limits improper-integrals
edited Dec 4 '18 at 23:51
MOMO
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
edited Dec 4 '18 at 23:51
MOMO
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
edited Dec 4 '18 at 23:51
MOMO
354110
edited Dec 4 '18 at 23:51
MOMO
354110
edited Dec 4 '18 at 23:51
MOMO
354110
354110
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
asked Dec 4 '18 at 21:37
NotEinsteinNotEinstein
1424
1424
add a comment |
add a comment |
1 Answer
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The claim is correct: Using the assumption show that
$$lim_{krightarrowinfty}int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_krightarrowinfty$, or equivalently
$$lim_{x,yrightarrowinfty}int_x^yf=0$$
Hence By Cauchy $lim_{xrightarrowinfty}int_0^xf$ converges. Similarly $lim_{xrightarrow-infty}int_x^0f$ converges so overall $f$ is integrable over $mathbb{R}$. Choosing $a_k=-k, b_k=k$ gives you the integral is $alpha$.
Your argument is wrong for the reason you have mentioned.
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
The claim is correct: Using the assumption show that
$$lim_{krightarrowinfty}int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_krightarrowinfty$, or equivalently
$$lim_{x,yrightarrowinfty}int_x^yf=0$$
Hence By Cauchy $lim_{xrightarrowinfty}int_0^xf$ converges. Similarly $lim_{xrightarrow-infty}int_x^0f$ converges so overall $f$ is integrable over $mathbb{R}$. Choosing $a_k=-k, b_k=k$ gives you the integral is $alpha$.
Your argument is wrong for the reason you have mentioned.
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
add a comment |
The claim is correct: Using the assumption show that
$$lim_{krightarrowinfty}int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_krightarrowinfty$, or equivalently
$$lim_{x,yrightarrowinfty}int_x^yf=0$$
Hence By Cauchy $lim_{xrightarrowinfty}int_0^xf$ converges. Similarly $lim_{xrightarrow-infty}int_x^0f$ converges so overall $f$ is integrable over $mathbb{R}$. Choosing $a_k=-k, b_k=k$ gives you the integral is $alpha$.
Your argument is wrong for the reason you have mentioned.
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
add a comment |
The claim is correct: Using the assumption show that
$$lim_{krightarrowinfty}int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_krightarrowinfty$, or equivalently
$$lim_{x,yrightarrowinfty}int_x^yf=0$$
Hence By Cauchy $lim_{xrightarrowinfty}int_0^xf$ converges. Similarly $lim_{xrightarrow-infty}int_x^0f$ converges so overall $f$ is integrable over $mathbb{R}$. Choosing $a_k=-k, b_k=k$ gives you the integral is $alpha$.
Your argument is wrong for the reason you have mentioned.
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
The claim is correct: Using the assumption show that
$$lim_{krightarrowinfty}int_{c_k}^{d_k}f=0$$
for any two sequences $c_k,d_krightarrowinfty$, or equivalently
$$lim_{x,yrightarrowinfty}int_x^yf=0$$
Hence By Cauchy $lim_{xrightarrowinfty}int_0^xf$ converges. Similarly $lim_{xrightarrow-infty}int_x^0f$ converges so overall $f$ is integrable over $mathbb{R}$. Choosing $a_k=-k, b_k=k$ gives you the integral is $alpha$.
Your argument is wrong for the reason you have mentioned.
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
answered Dec 4 '18 at 23:39
MOMOMOMO
354110
354110
add a comment |
add a comment |
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